An 80.0-kg skydiver jumps out of a balloon at an altitude of 1.00 × 10 3 m and opens the parachute at an altitude of 200.0 m. (a) Assuming that the total retarding force on the diver is constant at 50.0 N with the parachute closed and constant at 3.60 × 10 3 N with the parachute open, what is the speed of the diver when he lands on the ground? (b) Do you think the skydiver will get hurt? Explain. (c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s? (d) How realistic is the assumption that the total retarding force is constant? Explain.

Question

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Answer 1

Textbook Question

Chapter 5, Problem 49P

An 80.0-kg skydiver jumps out of a balloon at an altitude of 1.00 × 10³ m and opens the parachute at an altitude of 200.0 m. (a) Assuming that the total retarding force on the diver is constant at 50.0 N with the parachute closed and constant at 3.60 × 10³ N with the parachute open, what is the speed of the diver when he lands on the ground? (b) Do you think the skydiver will get hurt? Explain. (c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s? (d) How realistic is the assumption that the total retarding force is constant? Explain.

(a)

Expert Solution

To determine

The speed of the diver when he lands on the ground.

Answer to Problem 49P

The speed of the diver when he lands on the ground is

$24.5 m/s$ .

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The parachute opens at an altitude of $200.0 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)d1+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d1$ is the distance that the diver travels with closed parachute
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate speed of the diver when he lands on the ground is,

$vf=2(ghi−F1d1+F2d2m)$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $8000 m$ for $d1$ , $3.60×103 N$ for $F2$ , $200.0 m$ for $d2$ and $80.0 kg$ for m to find the speed of the diver,

$vf=2[(9.8 m/s2)(1.0×103 m)−(50.0 N)(8000 m)+(3.60×103 N)(200.0 m)80.0 kg]=24.5 m/s$

Thus, the speed of the diver when he lands on the ground is $24.5 m/s$ .

Conclusion:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

(b)

Expert Solution

To determine

Whether the skydiver will get hurt.

Answer to Problem 49P

The landing speed

$24.5 m/s$ is enough to get hurt.

Explanation of Solution

Given Info:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

The speed of the skydiver is more than enough to get hurt. At the landing speed of $24.5 m/s$ , he can’t balance in his feet. He will fall down and will get hurt.

Conclusion:

The landing speed $24.5 m/s$ is enough to get hurt.

(c)

Expert Solution

To determine

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$

Answer to Problem 49P

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$ is

$206 m$ .

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

The speed of the diver at the ground is $5.0 m/s$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Consider the height from which the parachute is a variable.

$d1=1000 m−d2$

$d1$ is the distance that the diver travels with closed parachute

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)(1000 m−d2)+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate the height with which the parachute should be opened is,

$d2=(mg)hi−(1000 m)f1−12mvf2F2−F1$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $3.60×103 N$ for $F2$ , $5.0 m/s$ for $vf$ and $80.0 kg$ for m to find the speed of the diver,

$d2=[(80.0 kg)(9.8 m/s2)]−(1.0×103 m)(50.0 N)−12(80.0 kg)(5.0 m/s)23.60×103 N−50.0 N=206 m$

Thus, the height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

Conclusion:

The height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

(d)

Expert Solution

To determine

how realistic is the assumption that the total retarding force is a constant.

Answer to Problem 49P

The retarding force will be depending on the speed of the diver in reality.

Explanation of Solution

In the actual case of sky diving, the air drag will be in accordance with the speed of the skydiver.

The weight of the skydiver is about $(80.0 kg)(9.8 m/s2) N$

Before the parachute open, the air drag will be less than the weight of the diver.

The air drag will be almost equal to the weight of the diver when he opens the chute and again before he touches down.

Conclusion:

The retarding force will be depending on the speed of the diver in reality.

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Answer 2

Textbook Question

Chapter 5, Problem 49P

An 80.0-kg skydiver jumps out of a balloon at an altitude of 1.00 × 10³ m and opens the parachute at an altitude of 200.0 m. (a) Assuming that the total retarding force on the diver is constant at 50.0 N with the parachute closed and constant at 3.60 × 10³ N with the parachute open, what is the speed of the diver when he lands on the ground? (b) Do you think the skydiver will get hurt? Explain. (c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s? (d) How realistic is the assumption that the total retarding force is constant? Explain.

(a)

Expert Solution

To determine

The speed of the diver when he lands on the ground.

Answer to Problem 49P

The speed of the diver when he lands on the ground is

$24.5 m/s$ .

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The parachute opens at an altitude of $200.0 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)d1+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d1$ is the distance that the diver travels with closed parachute
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate speed of the diver when he lands on the ground is,

$vf=2(ghi−F1d1+F2d2m)$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $8000 m$ for $d1$ , $3.60×103 N$ for $F2$ , $200.0 m$ for $d2$ and $80.0 kg$ for m to find the speed of the diver,

$vf=2[(9.8 m/s2)(1.0×103 m)−(50.0 N)(8000 m)+(3.60×103 N)(200.0 m)80.0 kg]=24.5 m/s$

Thus, the speed of the diver when he lands on the ground is $24.5 m/s$ .

Conclusion:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

(b)

Expert Solution

To determine

Whether the skydiver will get hurt.

Answer to Problem 49P

The landing speed

$24.5 m/s$ is enough to get hurt.

Explanation of Solution

Given Info:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

The speed of the skydiver is more than enough to get hurt. At the landing speed of $24.5 m/s$ , he can’t balance in his feet. He will fall down and will get hurt.

Conclusion:

The landing speed $24.5 m/s$ is enough to get hurt.

(c)

Expert Solution

To determine

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$

Answer to Problem 49P

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$ is

$206 m$ .

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

The speed of the diver at the ground is $5.0 m/s$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Consider the height from which the parachute is a variable.

$d1=1000 m−d2$

$d1$ is the distance that the diver travels with closed parachute

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)(1000 m−d2)+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate the height with which the parachute should be opened is,

$d2=(mg)hi−(1000 m)f1−12mvf2F2−F1$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $3.60×103 N$ for $F2$ , $5.0 m/s$ for $vf$ and $80.0 kg$ for m to find the speed of the diver,

$d2=[(80.0 kg)(9.8 m/s2)]−(1.0×103 m)(50.0 N)−12(80.0 kg)(5.0 m/s)23.60×103 N−50.0 N=206 m$

Thus, the height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

Conclusion:

The height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

(d)

Expert Solution

To determine

how realistic is the assumption that the total retarding force is a constant.

Answer to Problem 49P

The retarding force will be depending on the speed of the diver in reality.

Explanation of Solution

In the actual case of sky diving, the air drag will be in accordance with the speed of the skydiver.

The weight of the skydiver is about $(80.0 kg)(9.8 m/s2) N$

Before the parachute open, the air drag will be less than the weight of the diver.

The air drag will be almost equal to the weight of the diver when he opens the chute and again before he touches down.

Conclusion:

The retarding force will be depending on the speed of the diver in reality.

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Answer 3

Textbook Question

Chapter 5, Problem 49P

An 80.0-kg skydiver jumps out of a balloon at an altitude of 1.00 × 10³ m and opens the parachute at an altitude of 200.0 m. (a) Assuming that the total retarding force on the diver is constant at 50.0 N with the parachute closed and constant at 3.60 × 10³ N with the parachute open, what is the speed of the diver when he lands on the ground? (b) Do you think the skydiver will get hurt? Explain. (c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s? (d) How realistic is the assumption that the total retarding force is constant? Explain.

(a)

Expert Solution

To determine

The speed of the diver when he lands on the ground.

Answer to Problem 49P

The speed of the diver when he lands on the ground is

$24.5 m/s$ .

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The parachute opens at an altitude of $200.0 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)d1+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d1$ is the distance that the diver travels with closed parachute
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate speed of the diver when he lands on the ground is,

$vf=2(ghi−F1d1+F2d2m)$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $8000 m$ for $d1$ , $3.60×103 N$ for $F2$ , $200.0 m$ for $d2$ and $80.0 kg$ for m to find the speed of the diver,

$vf=2[(9.8 m/s2)(1.0×103 m)−(50.0 N)(8000 m)+(3.60×103 N)(200.0 m)80.0 kg]=24.5 m/s$

Thus, the speed of the diver when he lands on the ground is $24.5 m/s$ .

Conclusion:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

(b)

Expert Solution

To determine

Whether the skydiver will get hurt.

Answer to Problem 49P

The landing speed

$24.5 m/s$ is enough to get hurt.

Explanation of Solution

Given Info:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

The speed of the skydiver is more than enough to get hurt. At the landing speed of $24.5 m/s$ , he can’t balance in his feet. He will fall down and will get hurt.

Conclusion:

The landing speed $24.5 m/s$ is enough to get hurt.

(c)

Expert Solution

To determine

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$

Answer to Problem 49P

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$ is

$206 m$ .

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

The speed of the diver at the ground is $5.0 m/s$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Consider the height from which the parachute is a variable.

$d1=1000 m−d2$

$d1$ is the distance that the diver travels with closed parachute

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)(1000 m−d2)+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate the height with which the parachute should be opened is,

$d2=(mg)hi−(1000 m)f1−12mvf2F2−F1$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $3.60×103 N$ for $F2$ , $5.0 m/s$ for $vf$ and $80.0 kg$ for m to find the speed of the diver,

$d2=[(80.0 kg)(9.8 m/s2)]−(1.0×103 m)(50.0 N)−12(80.0 kg)(5.0 m/s)23.60×103 N−50.0 N=206 m$

Thus, the height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

Conclusion:

The height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

(d)

Expert Solution

To determine

how realistic is the assumption that the total retarding force is a constant.

Answer to Problem 49P

The retarding force will be depending on the speed of the diver in reality.

Explanation of Solution

In the actual case of sky diving, the air drag will be in accordance with the speed of the skydiver.

The weight of the skydiver is about $(80.0 kg)(9.8 m/s2) N$

Before the parachute open, the air drag will be less than the weight of the diver.

The air drag will be almost equal to the weight of the diver when he opens the chute and again before he touches down.

Conclusion:

The retarding force will be depending on the speed of the diver in reality.

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Answer 4

Textbook Question

Chapter 5, Problem 49P

An 80.0-kg skydiver jumps out of a balloon at an altitude of 1.00 × 10³ m and opens the parachute at an altitude of 200.0 m. (a) Assuming that the total retarding force on the diver is constant at 50.0 N with the parachute closed and constant at 3.60 × 10³ N with the parachute open, what is the speed of the diver when he lands on the ground? (b) Do you think the skydiver will get hurt? Explain. (c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s? (d) How realistic is the assumption that the total retarding force is constant? Explain.

(a)

Expert Solution

To determine

The speed of the diver when he lands on the ground.

Answer to Problem 49P

The speed of the diver when he lands on the ground is

$24.5 m/s$ .

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The parachute opens at an altitude of $200.0 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)d1+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d1$ is the distance that the diver travels with closed parachute
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate speed of the diver when he lands on the ground is,

$vf=2(ghi−F1d1+F2d2m)$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $8000 m$ for $d1$ , $3.60×103 N$ for $F2$ , $200.0 m$ for $d2$ and $80.0 kg$ for m to find the speed of the diver,

$vf=2[(9.8 m/s2)(1.0×103 m)−(50.0 N)(8000 m)+(3.60×103 N)(200.0 m)80.0 kg]=24.5 m/s$

Thus, the speed of the diver when he lands on the ground is $24.5 m/s$ .

Conclusion:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

(b)

Expert Solution

To determine

Whether the skydiver will get hurt.

Answer to Problem 49P

The landing speed

$24.5 m/s$ is enough to get hurt.

Explanation of Solution

Given Info:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

The speed of the skydiver is more than enough to get hurt. At the landing speed of $24.5 m/s$ , he can’t balance in his feet. He will fall down and will get hurt.

Conclusion:

The landing speed $24.5 m/s$ is enough to get hurt.

(c)

Expert Solution

To determine

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$

Answer to Problem 49P

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$ is

$206 m$ .

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

The speed of the diver at the ground is $5.0 m/s$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Consider the height from which the parachute is a variable.

$d1=1000 m−d2$

$d1$ is the distance that the diver travels with closed parachute

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)(1000 m−d2)+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate the height with which the parachute should be opened is,

$d2=(mg)hi−(1000 m)f1−12mvf2F2−F1$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $3.60×103 N$ for $F2$ , $5.0 m/s$ for $vf$ and $80.0 kg$ for m to find the speed of the diver,

$d2=[(80.0 kg)(9.8 m/s2)]−(1.0×103 m)(50.0 N)−12(80.0 kg)(5.0 m/s)23.60×103 N−50.0 N=206 m$

Thus, the height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

Conclusion:

The height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

(d)

Expert Solution

To determine

how realistic is the assumption that the total retarding force is a constant.

Answer to Problem 49P

The retarding force will be depending on the speed of the diver in reality.

Explanation of Solution

In the actual case of sky diving, the air drag will be in accordance with the speed of the skydiver.

The weight of the skydiver is about $(80.0 kg)(9.8 m/s2) N$

Before the parachute open, the air drag will be less than the weight of the diver.

The air drag will be almost equal to the weight of the diver when he opens the chute and again before he touches down.

Conclusion:

The retarding force will be depending on the speed of the diver in reality.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The speed of the diver when he lands on the ground.

Answer to Problem 49P

The speed of the diver when he lands on the ground is

$24.5 m/s$ .

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The parachute opens at an altitude of $200.0 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)d1+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d1$ is the distance that the diver travels with closed parachute
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate speed of the diver when he lands on the ground is,

$vf=2(ghi−F1d1+F2d2m)$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $8000 m$ for $d1$ , $3.60×103 N$ for $F2$ , $200.0 m$ for $d2$ and $80.0 kg$ for m to find the speed of the diver,

$vf=2[(9.8 m/s2)(1.0×103 m)−(50.0 N)(8000 m)+(3.60×103 N)(200.0 m)80.0 kg]=24.5 m/s$

Thus, the speed of the diver when he lands on the ground is $24.5 m/s$ .

Conclusion:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

(b)

Expert Solution

To determine

Whether the skydiver will get hurt.

Answer to Problem 49P

The landing speed

$24.5 m/s$ is enough to get hurt.

Explanation of Solution

Given Info:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

The speed of the skydiver is more than enough to get hurt. At the landing speed of $24.5 m/s$ , he can’t balance in his feet. He will fall down and will get hurt.

Conclusion:

The landing speed $24.5 m/s$ is enough to get hurt.

(c)

Expert Solution

To determine

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$

Answer to Problem 49P

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$ is

$206 m$ .

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

The speed of the diver at the ground is $5.0 m/s$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Consider the height from which the parachute is a variable.

$d1=1000 m−d2$

$d1$ is the distance that the diver travels with closed parachute

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)(1000 m−d2)+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate the height with which the parachute should be opened is,

$d2=(mg)hi−(1000 m)f1−12mvf2F2−F1$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $3.60×103 N$ for $F2$ , $5.0 m/s$ for $vf$ and $80.0 kg$ for m to find the speed of the diver,

$d2=[(80.0 kg)(9.8 m/s2)]−(1.0×103 m)(50.0 N)−12(80.0 kg)(5.0 m/s)23.60×103 N−50.0 N=206 m$

Thus, the height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

Conclusion:

The height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

(d)

Expert Solution

To determine

how realistic is the assumption that the total retarding force is a constant.

Answer to Problem 49P

The retarding force will be depending on the speed of the diver in reality.

Explanation of Solution

In the actual case of sky diving, the air drag will be in accordance with the speed of the skydiver.

The weight of the skydiver is about $(80.0 kg)(9.8 m/s2) N$

Before the parachute open, the air drag will be less than the weight of the diver.

The air drag will be almost equal to the weight of the diver when he opens the chute and again before he touches down.

Conclusion:

The retarding force will be depending on the speed of the diver in reality.

Answer 8

(a)

Expert Solution

To determine

The speed of the diver when he lands on the ground.

Answer to Problem 49P

The speed of the diver when he lands on the ground is

$24.5 m/s$ .

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The parachute opens at an altitude of $200.0 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)d1+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d1$ is the distance that the diver travels with closed parachute
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate speed of the diver when he lands on the ground is,

$vf=2(ghi−F1d1+F2d2m)$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $8000 m$ for $d1$ , $3.60×103 N$ for $F2$ , $200.0 m$ for $d2$ and $80.0 kg$ for m to find the speed of the diver,

$vf=2[(9.8 m/s2)(1.0×103 m)−(50.0 N)(8000 m)+(3.60×103 N)(200.0 m)80.0 kg]=24.5 m/s$

Thus, the speed of the diver when he lands on the ground is $24.5 m/s$ .

Conclusion:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The speed of the diver when he lands on the ground.

Answer 13

Answer to Problem 49P

The speed of the diver when he lands on the ground is

$24.5 m/s$ .

Answer 14

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The parachute opens at an altitude of $200.0 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)d1+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d1$ is the distance that the diver travels with closed parachute
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate speed of the diver when he lands on the ground is,

$vf=2(ghi−F1d1+F2d2m)$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $8000 m$ for $d1$ , $3.60×103 N$ for $F2$ , $200.0 m$ for $d2$ and $80.0 kg$ for m to find the speed of the diver,

$vf=2[(9.8 m/s2)(1.0×103 m)−(50.0 N)(8000 m)+(3.60×103 N)(200.0 m)80.0 kg]=24.5 m/s$

Thus, the speed of the diver when he lands on the ground is $24.5 m/s$ .

Conclusion:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

Answer 15

(b)

Expert Solution

To determine

Whether the skydiver will get hurt.

Answer to Problem 49P

The landing speed

$24.5 m/s$ is enough to get hurt.

Explanation of Solution

Given Info:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

The speed of the skydiver is more than enough to get hurt. At the landing speed of $24.5 m/s$ , he can’t balance in his feet. He will fall down and will get hurt.

Conclusion:

The landing speed $24.5 m/s$ is enough to get hurt.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Whether the skydiver will get hurt.

Answer 20

Answer to Problem 49P

The landing speed

$24.5 m/s$ is enough to get hurt.

Answer 21

Explanation of Solution

Given Info:

The speed of the diver when he lands on the ground is $24.5 m/s$ .

The speed of the skydiver is more than enough to get hurt. At the landing speed of $24.5 m/s$ , he can’t balance in his feet. He will fall down and will get hurt.

Conclusion:

The landing speed $24.5 m/s$ is enough to get hurt.

Answer 22

(c)

Expert Solution

To determine

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$

Answer to Problem 49P

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$ is

$206 m$ .

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

The speed of the diver at the ground is $5.0 m/s$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Consider the height from which the parachute is a variable.

$d1=1000 m−d2$

$d1$ is the distance that the diver travels with closed parachute

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)(1000 m−d2)+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate the height with which the parachute should be opened is,

$d2=(mg)hi−(1000 m)f1−12mvf2F2−F1$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $3.60×103 N$ for $F2$ , $5.0 m/s$ for $vf$ and $80.0 kg$ for m to find the speed of the diver,

$d2=[(80.0 kg)(9.8 m/s2)]−(1.0×103 m)(50.0 N)−12(80.0 kg)(5.0 m/s)23.60×103 N−50.0 N=206 m$

Thus, the height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

Conclusion:

The height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$

Answer 27

Answer to Problem 49P

The height with which the parachute should be opened if he needs a landing speed of

$5.0 m/s$ is

$206 m$ .

Answer 28

Explanation of Solution

Given Info:

The mass of the sky diver is $80.0 kg$ .

The height of the balloon from which the sky diver jumps is $1.0×103 m$ .

The total retarding force when the parachute is open is $50.0 N$ .

The total retarding force when the parachute is closed is $3.60×103 N$ .

The speed of the diver at the ground is $5.0 m/s$ .

According to the Work-Energy theorem in the entire trip,

$Wnc=(KE+PEg)f−(KE+PEg)i$

Consider the height from which the parachute is a variable.

$d1=1000 m−d2$

$d1$ is the distance that the diver travels with closed parachute

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

$(F1cos180°)(1000 m−d2)+(F2cos180°)d2=(12mvf2+0)−(0+mghi)$

$F1$ is the total retarding force when the parachute is open
$F2$ is the total retarding force when the parachute is closed
$d2$ is the height from which the parachute is open
m is the mass of diver
$hi$ is the initial height of the diver
$vf$ is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate the height with which the parachute should be opened is,

$d2=(mg)hi−(1000 m)f1−12mvf2F2−F1$

Substitute $9.8 m/s2$ for $g$ , $1.0×103 m$ for $hi$ , $50.0 N$ for $F1$ , $3.60×103 N$ for $F2$ , $5.0 m/s$ for $vf$ and $80.0 kg$ for m to find the speed of the diver,

$d2=[(80.0 kg)(9.8 m/s2)]−(1.0×103 m)(50.0 N)−12(80.0 kg)(5.0 m/s)23.60×103 N−50.0 N=206 m$

Thus, the height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

Conclusion:

The height with which the parachute should be opened if he needs a landing speed of $5.0 m/s$ is $206 m$ .

Answer 29

(d)

Expert Solution

To determine

how realistic is the assumption that the total retarding force is a constant.

Answer to Problem 49P

The retarding force will be depending on the speed of the diver in reality.

Explanation of Solution

In the actual case of sky diving, the air drag will be in accordance with the speed of the skydiver.

The weight of the skydiver is about $(80.0 kg)(9.8 m/s2) N$

Before the parachute open, the air drag will be less than the weight of the diver.

The air drag will be almost equal to the weight of the diver when he opens the chute and again before he touches down.

Conclusion:

The retarding force will be depending on the speed of the diver in reality.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

how realistic is the assumption that the total retarding force is a constant.

Answer 34

Answer to Problem 49P

The retarding force will be depending on the speed of the diver in reality.

Answer 35

Explanation of Solution

In the actual case of sky diving, the air drag will be in accordance with the speed of the skydiver.

The weight of the skydiver is about $(80.0 kg)(9.8 m/s2) N$

Before the parachute open, the air drag will be less than the weight of the diver.

The air drag will be almost equal to the weight of the diver when he opens the chute and again before he touches down.

Conclusion:

The retarding force will be depending on the speed of the diver in reality.

Answer 36

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