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

Question

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Answer 1

Question

Chapter 7, Problem 24P

(a)

To determine

To determine: The speed of the sky diver when he lands on the ground.

(a)

Expert Solution

Answer to Problem 24P

Answer: The speed of the sky diver when he lands on the ground is $24.9 m/s$ .

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Formula to calculate the speed of the sky diver by conservation of energy is,

$K1+U1+ΔE=K2+U2$ (I)

$K1$ is the kinetic energy of the sky diver before jump.

$K2$ is the kinetic energy of the sky diver when he lands at ground.

$U1$ is the potential energy of the sky diver before jump.

$U2$ is the potential energy the sky diver to reached the ground.

$ΔE$ is the total mechanical energy.

The initial kinetic energy of the sky diver is zero because sky diver is initially at rest and final potential energy of the sky diver is zero because the distance is zero.

Formula to calculate the potential energy of the sky diver before jump is,

$U1=mgh$

$h$ is the distance between ground and sky diver jump.

$m$ is the mass of the sky diver.

$g$ is the acceleration due to gravity.

Formula to calculate the total mechanical energy of the system is,

$ΔE=F1Δx1+F2Δx2$

$F1$ is the force acting on the sky diver $Δx1$ when the parachute is closed.

$F2$ is the force acting on the sky diver when the parachute is open.

$Δx1$ is the distance covered by the sky diver when the parachute is closed.

$Δx2$ is the distance covered by the sky diver when the parachute is open.

Formula to calculate the kinetic energy of the sky diver when he lands on ground is,

$K2=12mv2$

$v$ is the velocity of the sky diver when he lands a ground.

Substitute $12mv2$ for $K2$ , $F1Δx1+F2Δx2$ for $ΔE$ , $mgh$ for $U1$ , $0$ for $U2$ and $0$ for $K1$ in equation (I).

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$ (II)

The force $F1$ and $F2$ is negative because direction is opposite to the weight of the sky diver.

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $800 m$ for $Δx1$ , $200 m$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ and $9.81 m/s2$ for $g$ to find $v$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×800 m+3600 N×200 m))=(1280.0 kg×v2+0)v=24.9 m/s$

Conclusion:

Therefore, the speed of the sky diver when he lands on the ground is $24.9 m/s$ .

(b)

To determine

To explain: Whether the sky diver will be injured or not.

(b)

Expert Solution

Answer to Problem 24P

Answer: Therefore, the sky diver will be injured because his speed is very high.

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Yes, the sky diver will be injured because speed of the sky diver is covered the distance approx $25 m$ in one second. This speed is very high as compared to normal speed so the force of impact is so high when he lands on the ground that he will be injured.

Conclusion:

Therefore, the sky diver will be injured because his speed is very high.

(c)

To determine

To determine: The height at which parachute should be open if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

(c)

Expert Solution

Answer to Problem 24P

Answer: The height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

From equation (II),

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$

Assume $Δx2=x$ so $Δx1=1000 m−x$ .

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $1000 m−x$ for $Δx1$ , $x$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ , $5.0 m/s$ for $v$ and $9.81 m/s2$ for $g$ to find $x$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×(1000 m−x)−3600 N×x))=(12×80.0 kg×(5.0 m/s)2+0)784800 N-m−50000 N-m+(50 N)x−(3600 N)x=1000 N-m(3350 N)x=733800 N-mx=219 m$

Conclusion:

Therefore, height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

(d)

To determine

To determine: The assumption that the total retarding force is constant.

(d)

Expert Solution

Answer to Problem 24P

Answer: Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

The assumption that total retarding force is constant is not realistic because the air density changes with the change in altitude. Retarding force is proportional to the density of the air so, with the change in the density of air, the retarding force also changes with the altitude.

Conclusion:

Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

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Answer 2

Question

Chapter 7, Problem 24P

(a)

To determine

To determine: The speed of the sky diver when he lands on the ground.

(a)

Expert Solution

Answer to Problem 24P

Answer: The speed of the sky diver when he lands on the ground is $24.9 m/s$ .

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Formula to calculate the speed of the sky diver by conservation of energy is,

$K1+U1+ΔE=K2+U2$ (I)

$K1$ is the kinetic energy of the sky diver before jump.

$K2$ is the kinetic energy of the sky diver when he lands at ground.

$U1$ is the potential energy of the sky diver before jump.

$U2$ is the potential energy the sky diver to reached the ground.

$ΔE$ is the total mechanical energy.

The initial kinetic energy of the sky diver is zero because sky diver is initially at rest and final potential energy of the sky diver is zero because the distance is zero.

Formula to calculate the potential energy of the sky diver before jump is,

$U1=mgh$

$h$ is the distance between ground and sky diver jump.

$m$ is the mass of the sky diver.

$g$ is the acceleration due to gravity.

Formula to calculate the total mechanical energy of the system is,

$ΔE=F1Δx1+F2Δx2$

$F1$ is the force acting on the sky diver $Δx1$ when the parachute is closed.

$F2$ is the force acting on the sky diver when the parachute is open.

$Δx1$ is the distance covered by the sky diver when the parachute is closed.

$Δx2$ is the distance covered by the sky diver when the parachute is open.

Formula to calculate the kinetic energy of the sky diver when he lands on ground is,

$K2=12mv2$

$v$ is the velocity of the sky diver when he lands a ground.

Substitute $12mv2$ for $K2$ , $F1Δx1+F2Δx2$ for $ΔE$ , $mgh$ for $U1$ , $0$ for $U2$ and $0$ for $K1$ in equation (I).

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$ (II)

The force $F1$ and $F2$ is negative because direction is opposite to the weight of the sky diver.

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $800 m$ for $Δx1$ , $200 m$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ and $9.81 m/s2$ for $g$ to find $v$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×800 m+3600 N×200 m))=(1280.0 kg×v2+0)v=24.9 m/s$

Conclusion:

Therefore, the speed of the sky diver when he lands on the ground is $24.9 m/s$ .

(b)

To determine

To explain: Whether the sky diver will be injured or not.

(b)

Expert Solution

Answer to Problem 24P

Answer: Therefore, the sky diver will be injured because his speed is very high.

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Yes, the sky diver will be injured because speed of the sky diver is covered the distance approx $25 m$ in one second. This speed is very high as compared to normal speed so the force of impact is so high when he lands on the ground that he will be injured.

Conclusion:

Therefore, the sky diver will be injured because his speed is very high.

(c)

To determine

To determine: The height at which parachute should be open if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

(c)

Expert Solution

Answer to Problem 24P

Answer: The height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

From equation (II),

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$

Assume $Δx2=x$ so $Δx1=1000 m−x$ .

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $1000 m−x$ for $Δx1$ , $x$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ , $5.0 m/s$ for $v$ and $9.81 m/s2$ for $g$ to find $x$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×(1000 m−x)−3600 N×x))=(12×80.0 kg×(5.0 m/s)2+0)784800 N-m−50000 N-m+(50 N)x−(3600 N)x=1000 N-m(3350 N)x=733800 N-mx=219 m$

Conclusion:

Therefore, height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

(d)

To determine

To determine: The assumption that the total retarding force is constant.

(d)

Expert Solution

Answer to Problem 24P

Answer: Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

The assumption that total retarding force is constant is not realistic because the air density changes with the change in altitude. Retarding force is proportional to the density of the air so, with the change in the density of air, the retarding force also changes with the altitude.

Conclusion:

Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

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Answer 3

Question

Chapter 7, Problem 24P

(a)

To determine

To determine: The speed of the sky diver when he lands on the ground.

(a)

Expert Solution

Answer to Problem 24P

Answer: The speed of the sky diver when he lands on the ground is $24.9 m/s$ .

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Formula to calculate the speed of the sky diver by conservation of energy is,

$K1+U1+ΔE=K2+U2$ (I)

$K1$ is the kinetic energy of the sky diver before jump.

$K2$ is the kinetic energy of the sky diver when he lands at ground.

$U1$ is the potential energy of the sky diver before jump.

$U2$ is the potential energy the sky diver to reached the ground.

$ΔE$ is the total mechanical energy.

The initial kinetic energy of the sky diver is zero because sky diver is initially at rest and final potential energy of the sky diver is zero because the distance is zero.

Formula to calculate the potential energy of the sky diver before jump is,

$U1=mgh$

$h$ is the distance between ground and sky diver jump.

$m$ is the mass of the sky diver.

$g$ is the acceleration due to gravity.

Formula to calculate the total mechanical energy of the system is,

$ΔE=F1Δx1+F2Δx2$

$F1$ is the force acting on the sky diver $Δx1$ when the parachute is closed.

$F2$ is the force acting on the sky diver when the parachute is open.

$Δx1$ is the distance covered by the sky diver when the parachute is closed.

$Δx2$ is the distance covered by the sky diver when the parachute is open.

Formula to calculate the kinetic energy of the sky diver when he lands on ground is,

$K2=12mv2$

$v$ is the velocity of the sky diver when he lands a ground.

Substitute $12mv2$ for $K2$ , $F1Δx1+F2Δx2$ for $ΔE$ , $mgh$ for $U1$ , $0$ for $U2$ and $0$ for $K1$ in equation (I).

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$ (II)

The force $F1$ and $F2$ is negative because direction is opposite to the weight of the sky diver.

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $800 m$ for $Δx1$ , $200 m$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ and $9.81 m/s2$ for $g$ to find $v$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×800 m+3600 N×200 m))=(1280.0 kg×v2+0)v=24.9 m/s$

Conclusion:

Therefore, the speed of the sky diver when he lands on the ground is $24.9 m/s$ .

(b)

To determine

To explain: Whether the sky diver will be injured or not.

(b)

Expert Solution

Answer to Problem 24P

Answer: Therefore, the sky diver will be injured because his speed is very high.

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Yes, the sky diver will be injured because speed of the sky diver is covered the distance approx $25 m$ in one second. This speed is very high as compared to normal speed so the force of impact is so high when he lands on the ground that he will be injured.

Conclusion:

Therefore, the sky diver will be injured because his speed is very high.

(c)

To determine

To determine: The height at which parachute should be open if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

(c)

Expert Solution

Answer to Problem 24P

Answer: The height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

From equation (II),

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$

Assume $Δx2=x$ so $Δx1=1000 m−x$ .

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $1000 m−x$ for $Δx1$ , $x$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ , $5.0 m/s$ for $v$ and $9.81 m/s2$ for $g$ to find $x$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×(1000 m−x)−3600 N×x))=(12×80.0 kg×(5.0 m/s)2+0)784800 N-m−50000 N-m+(50 N)x−(3600 N)x=1000 N-m(3350 N)x=733800 N-mx=219 m$

Conclusion:

Therefore, height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

(d)

To determine

To determine: The assumption that the total retarding force is constant.

(d)

Expert Solution

Answer to Problem 24P

Answer: Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

The assumption that total retarding force is constant is not realistic because the air density changes with the change in altitude. Retarding force is proportional to the density of the air so, with the change in the density of air, the retarding force also changes with the altitude.

Conclusion:

Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 7, Problem 24P

(a)

To determine

To determine: The speed of the sky diver when he lands on the ground.

(a)

Expert Solution

Answer to Problem 24P

Answer: The speed of the sky diver when he lands on the ground is $24.9 m/s$ .

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Formula to calculate the speed of the sky diver by conservation of energy is,

$K1+U1+ΔE=K2+U2$ (I)

$K1$ is the kinetic energy of the sky diver before jump.

$K2$ is the kinetic energy of the sky diver when he lands at ground.

$U1$ is the potential energy of the sky diver before jump.

$U2$ is the potential energy the sky diver to reached the ground.

$ΔE$ is the total mechanical energy.

The initial kinetic energy of the sky diver is zero because sky diver is initially at rest and final potential energy of the sky diver is zero because the distance is zero.

Formula to calculate the potential energy of the sky diver before jump is,

$U1=mgh$

$h$ is the distance between ground and sky diver jump.

$m$ is the mass of the sky diver.

$g$ is the acceleration due to gravity.

Formula to calculate the total mechanical energy of the system is,

$ΔE=F1Δx1+F2Δx2$

$F1$ is the force acting on the sky diver $Δx1$ when the parachute is closed.

$F2$ is the force acting on the sky diver when the parachute is open.

$Δx1$ is the distance covered by the sky diver when the parachute is closed.

$Δx2$ is the distance covered by the sky diver when the parachute is open.

Formula to calculate the kinetic energy of the sky diver when he lands on ground is,

$K2=12mv2$

$v$ is the velocity of the sky diver when he lands a ground.

Substitute $12mv2$ for $K2$ , $F1Δx1+F2Δx2$ for $ΔE$ , $mgh$ for $U1$ , $0$ for $U2$ and $0$ for $K1$ in equation (I).

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$ (II)

The force $F1$ and $F2$ is negative because direction is opposite to the weight of the sky diver.

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $800 m$ for $Δx1$ , $200 m$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ and $9.81 m/s2$ for $g$ to find $v$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×800 m+3600 N×200 m))=(1280.0 kg×v2+0)v=24.9 m/s$

Conclusion:

Therefore, the speed of the sky diver when he lands on the ground is $24.9 m/s$ .

(b)

To determine

To explain: Whether the sky diver will be injured or not.

(b)

Expert Solution

Answer to Problem 24P

Answer: Therefore, the sky diver will be injured because his speed is very high.

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Yes, the sky diver will be injured because speed of the sky diver is covered the distance approx $25 m$ in one second. This speed is very high as compared to normal speed so the force of impact is so high when he lands on the ground that he will be injured.

Conclusion:

Therefore, the sky diver will be injured because his speed is very high.

(c)

To determine

To determine: The height at which parachute should be open if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

(c)

Expert Solution

Answer to Problem 24P

Answer: The height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

From equation (II),

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$

Assume $Δx2=x$ so $Δx1=1000 m−x$ .

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $1000 m−x$ for $Δx1$ , $x$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ , $5.0 m/s$ for $v$ and $9.81 m/s2$ for $g$ to find $x$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×(1000 m−x)−3600 N×x))=(12×80.0 kg×(5.0 m/s)2+0)784800 N-m−50000 N-m+(50 N)x−(3600 N)x=1000 N-m(3350 N)x=733800 N-mx=219 m$

Conclusion:

Therefore, height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

(d)

To determine

To determine: The assumption that the total retarding force is constant.

(d)

Expert Solution

Answer to Problem 24P

Answer: Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

The assumption that total retarding force is constant is not realistic because the air density changes with the change in altitude. Retarding force is proportional to the density of the air so, with the change in the density of air, the retarding force also changes with the altitude.

Conclusion:

Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 7, Problem 24P

(a)

To determine

To determine: The speed of the sky diver when he lands on the ground.

(a)

Expert Solution

Answer to Problem 24P

Answer: The speed of the sky diver when he lands on the ground is $24.9 m/s$ .

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Formula to calculate the speed of the sky diver by conservation of energy is,

$K1+U1+ΔE=K2+U2$ (I)

$K1$ is the kinetic energy of the sky diver before jump.

$K2$ is the kinetic energy of the sky diver when he lands at ground.

$U1$ is the potential energy of the sky diver before jump.

$U2$ is the potential energy the sky diver to reached the ground.

$ΔE$ is the total mechanical energy.

The initial kinetic energy of the sky diver is zero because sky diver is initially at rest and final potential energy of the sky diver is zero because the distance is zero.

Formula to calculate the potential energy of the sky diver before jump is,

$U1=mgh$

$h$ is the distance between ground and sky diver jump.

$m$ is the mass of the sky diver.

$g$ is the acceleration due to gravity.

Formula to calculate the total mechanical energy of the system is,

$ΔE=F1Δx1+F2Δx2$

$F1$ is the force acting on the sky diver $Δx1$ when the parachute is closed.

$F2$ is the force acting on the sky diver when the parachute is open.

$Δx1$ is the distance covered by the sky diver when the parachute is closed.

$Δx2$ is the distance covered by the sky diver when the parachute is open.

Formula to calculate the kinetic energy of the sky diver when he lands on ground is,

$K2=12mv2$

$v$ is the velocity of the sky diver when he lands a ground.

Substitute $12mv2$ for $K2$ , $F1Δx1+F2Δx2$ for $ΔE$ , $mgh$ for $U1$ , $0$ for $U2$ and $0$ for $K1$ in equation (I).

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$ (II)

The force $F1$ and $F2$ is negative because direction is opposite to the weight of the sky diver.

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $800 m$ for $Δx1$ , $200 m$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ and $9.81 m/s2$ for $g$ to find $v$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×800 m+3600 N×200 m))=(1280.0 kg×v2+0)v=24.9 m/s$

Conclusion:

Therefore, the speed of the sky diver when he lands on the ground is $24.9 m/s$ .

(b)

To determine

To explain: Whether the sky diver will be injured or not.

(b)

Expert Solution

Answer to Problem 24P

Answer: Therefore, the sky diver will be injured because his speed is very high.

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Yes, the sky diver will be injured because speed of the sky diver is covered the distance approx $25 m$ in one second. This speed is very high as compared to normal speed so the force of impact is so high when he lands on the ground that he will be injured.

Conclusion:

Therefore, the sky diver will be injured because his speed is very high.

(c)

To determine

To determine: The height at which parachute should be open if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

(c)

Expert Solution

Answer to Problem 24P

Answer: The height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

From equation (II),

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$

Assume $Δx2=x$ so $Δx1=1000 m−x$ .

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $1000 m−x$ for $Δx1$ , $x$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ , $5.0 m/s$ for $v$ and $9.81 m/s2$ for $g$ to find $x$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×(1000 m−x)−3600 N×x))=(12×80.0 kg×(5.0 m/s)2+0)784800 N-m−50000 N-m+(50 N)x−(3600 N)x=1000 N-m(3350 N)x=733800 N-mx=219 m$

Conclusion:

Therefore, height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

(d)

To determine

To determine: The assumption that the total retarding force is constant.

(d)

Expert Solution

Answer to Problem 24P

Answer: Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

The assumption that total retarding force is constant is not realistic because the air density changes with the change in altitude. Retarding force is proportional to the density of the air so, with the change in the density of air, the retarding force also changes with the altitude.

Conclusion:

Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

Answer 6

Chapter 7, Problem 24P

(a)

To determine

To determine: The speed of the sky diver when he lands on the ground.

(a)

Expert Solution

Answer to Problem 24P

Answer: The speed of the sky diver when he lands on the ground is $24.9 m/s$ .

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Formula to calculate the speed of the sky diver by conservation of energy is,

$K1+U1+ΔE=K2+U2$ (I)

$K1$ is the kinetic energy of the sky diver before jump.

$K2$ is the kinetic energy of the sky diver when he lands at ground.

$U1$ is the potential energy of the sky diver before jump.

$U2$ is the potential energy the sky diver to reached the ground.

$ΔE$ is the total mechanical energy.

The initial kinetic energy of the sky diver is zero because sky diver is initially at rest and final potential energy of the sky diver is zero because the distance is zero.

Formula to calculate the potential energy of the sky diver before jump is,

$U1=mgh$

$h$ is the distance between ground and sky diver jump.

$m$ is the mass of the sky diver.

$g$ is the acceleration due to gravity.

Formula to calculate the total mechanical energy of the system is,

$ΔE=F1Δx1+F2Δx2$

$F1$ is the force acting on the sky diver $Δx1$ when the parachute is closed.

$F2$ is the force acting on the sky diver when the parachute is open.

$Δx1$ is the distance covered by the sky diver when the parachute is closed.

$Δx2$ is the distance covered by the sky diver when the parachute is open.

Formula to calculate the kinetic energy of the sky diver when he lands on ground is,

$K2=12mv2$

$v$ is the velocity of the sky diver when he lands a ground.

Substitute $12mv2$ for $K2$ , $F1Δx1+F2Δx2$ for $ΔE$ , $mgh$ for $U1$ , $0$ for $U2$ and $0$ for $K1$ in equation (I).

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$ (II)

The force $F1$ and $F2$ is negative because direction is opposite to the weight of the sky diver.

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $800 m$ for $Δx1$ , $200 m$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ and $9.81 m/s2$ for $g$ to find $v$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×800 m+3600 N×200 m))=(1280.0 kg×v2+0)v=24.9 m/s$

Conclusion:

Therefore, the speed of the sky diver when he lands on the ground is $24.9 m/s$ .

Answer 7

Chapter 7, Problem 24P

(a)

To determine

To determine: The speed of the sky diver when he lands on the ground.

Answer 8

(a)

Expert Solution

Answer to Problem 24P

Answer: The speed of the sky diver when he lands on the ground is $24.9 m/s$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Formula to calculate the speed of the sky diver by conservation of energy is,

$K1+U1+ΔE=K2+U2$ (I)

$K1$ is the kinetic energy of the sky diver before jump.

$K2$ is the kinetic energy of the sky diver when he lands at ground.

$U1$ is the potential energy of the sky diver before jump.

$U2$ is the potential energy the sky diver to reached the ground.

$ΔE$ is the total mechanical energy.

The initial kinetic energy of the sky diver is zero because sky diver is initially at rest and final potential energy of the sky diver is zero because the distance is zero.

Formula to calculate the potential energy of the sky diver before jump is,

$U1=mgh$

$h$ is the distance between ground and sky diver jump.

$m$ is the mass of the sky diver.

$g$ is the acceleration due to gravity.

Formula to calculate the total mechanical energy of the system is,

$ΔE=F1Δx1+F2Δx2$

$F1$ is the force acting on the sky diver $Δx1$ when the parachute is closed.

$F2$ is the force acting on the sky diver when the parachute is open.

$Δx1$ is the distance covered by the sky diver when the parachute is closed.

$Δx2$ is the distance covered by the sky diver when the parachute is open.

Formula to calculate the kinetic energy of the sky diver when he lands on ground is,

$K2=12mv2$

$v$ is the velocity of the sky diver when he lands a ground.

Substitute $12mv2$ for $K2$ , $F1Δx1+F2Δx2$ for $ΔE$ , $mgh$ for $U1$ , $0$ for $U2$ and $0$ for $K1$ in equation (I).

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$ (II)

The force $F1$ and $F2$ is negative because direction is opposite to the weight of the sky diver.

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $800 m$ for $Δx1$ , $200 m$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ and $9.81 m/s2$ for $g$ to find $v$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×800 m+3600 N×200 m))=(1280.0 kg×v2+0)v=24.9 m/s$

Conclusion:

Therefore, the speed of the sky diver when he lands on the ground is $24.9 m/s$ .

Answer 13

(b)

To determine

To explain: Whether the sky diver will be injured or not.

(b)

Expert Solution

Answer to Problem 24P

Answer: Therefore, the sky diver will be injured because his speed is very high.

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Yes, the sky diver will be injured because speed of the sky diver is covered the distance approx $25 m$ in one second. This speed is very high as compared to normal speed so the force of impact is so high when he lands on the ground that he will be injured.

Conclusion:

Therefore, the sky diver will be injured because his speed is very high.

Answer 14

(b)

To determine

To explain: Whether the sky diver will be injured or not.

Answer 15

(b)

Expert Solution

Answer to Problem 24P

Answer: Therefore, the sky diver will be injured because his speed is very high.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ . Total retarding force on the skydiver with the parachute closed is $50.0 N$ and with the parachute open is $3600 N$ .

Yes, the sky diver will be injured because speed of the sky diver is covered the distance approx $25 m$ in one second. This speed is very high as compared to normal speed so the force of impact is so high when he lands on the ground that he will be injured.

Conclusion:

Therefore, the sky diver will be injured because his speed is very high.

Answer 20

(c)

To determine

To determine: The height at which parachute should be open if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

(c)

Expert Solution

Answer to Problem 24P

Answer: The height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

From equation (II),

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$

Assume $Δx2=x$ so $Δx1=1000 m−x$ .

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $1000 m−x$ for $Δx1$ , $x$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ , $5.0 m/s$ for $v$ and $9.81 m/s2$ for $g$ to find $x$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×(1000 m−x)−3600 N×x))=(12×80.0 kg×(5.0 m/s)2+0)784800 N-m−50000 N-m+(50 N)x−(3600 N)x=1000 N-m(3350 N)x=733800 N-mx=219 m$

Conclusion:

Therefore, height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

Answer 21

(c)

To determine

To determine: The height at which parachute should be open if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

Answer 22

(c)

Expert Solution

Answer to Problem 24P

Answer: The height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

From equation (II),

$(0+mgh−(F1Δx1+F2Δx2))=(12mv2+0)$

Assume $Δx2=x$ so $Δx1=1000 m−x$ .

Substitute $80.0 kg$ for $m$ , $1000 m$ for $h$ , $1000 m−x$ for $Δx1$ , $x$ for $Δx2$ , $50 N$ for $F1$ , $3600 N$ for $F2$ , $5.0 m/s$ for $v$ and $9.81 m/s2$ for $g$ to find $x$ .

$(0+80.0 kg×9.81 m/s2×1000 m−(50 N×(1000 m−x)−3600 N×x))=(12×80.0 kg×(5.0 m/s)2+0)784800 N-m−50000 N-m+(50 N)x−(3600 N)x=1000 N-m(3350 N)x=733800 N-mx=219 m$

Conclusion:

Therefore, height parachute should be open is $219 m$ if the final speed of the sky diver when he hits the ground is $5.00 m/s$ .

Answer 27

(d)

To determine

To determine: The assumption that the total retarding force is constant.

(d)

Expert Solution

Answer to Problem 24P

Answer: Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

The assumption that total retarding force is constant is not realistic because the air density changes with the change in altitude. Retarding force is proportional to the density of the air so, with the change in the density of air, the retarding force also changes with the altitude.

Conclusion:

Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

Answer 28

(d)

To determine

To determine: The assumption that the total retarding force is constant.

Answer 29

(d)

Expert Solution

Answer to Problem 24P

Answer: Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Explanation:

Given information:

An $80.0 kg$ sky diver jumps out of a balloon at an altitude of $1000 m$ and opens his parachute at an altitude of $200 m$ .

The assumption that total retarding force is constant is not realistic because the air density changes with the change in altitude. Retarding force is proportional to the density of the air so, with the change in the density of air, the retarding force also changes with the altitude.

Conclusion:

Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.