PHYSICS 1250 PACKAGE >CI<
PHYSICS 1250 PACKAGE >CI<
9th Edition
ISBN: 9781305000988
Author: SERWAY
Publisher: CENGAGE LEARNING (CUSTOM)
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Chapter 4, Problem 4.56AP

A ball is thrown with an initial speed υi at an angle θi with the horizontal. The horizontal range of the ball is R. and the ball reaches a maximum height R/6. In terms of R and g, find (a) the time interval during which the ball is in motion, (b) the ball’s speed at the peak of its path, (c) the initial vertical component of its velocity, (d) its initial speed, and (e) the angle θi, (f) Suppose the ball is thrown at the same initial speed found in (d) but at the angle appropriate for reach­ing the greatest height that it can. Find this height. (g) Suppose the ball is thrown at the same initial speed but at the angle for greatest possible range. Find this maximum horizontal range.

(a)

Expert Solution
Check Mark
To determine

The time interval during which the ball is in motion.

Answer to Problem 4.56AP

The time interval during which the ball is in motion is 4R3g .

Explanation of Solution

Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi , the horizontal range of the ball is R and the maximum height covered by the ball is R6

The motion of the ball follows the parabolic path and the ball is said to projectile, the motion of the ball is shown in the Figure below.

PHYSICS 1250 PACKAGE >CI<, Chapter 4, Problem 4.56AP

Figure (1)

The formula to calculate the maximum height reached by the projectile is,

H=(vi)2sin2θ2g

Here,

g is the acceleration due to gravity.

H is the maximum height reached by the ball.

Rearrange the above equation.

(vi)2sin2θ=2gHvisinθ=2gH

Substitute R6 for H in the above equation.

visinθ=2gR6=gR3

Thus, the vertical component of the initial velocity is gR3 .

The formula to calculate the time taken by the ball to reach the ground is,

t=2visinθg

Here,

t is the time taken by the ball to reach the ground.

Substitute gR3 for visinθ in the above equation.

t=2gR3g=4R3g

Conclusion:

Therefore, the time interval during which the ball is in motion is 4R3g .

(b)

Expert Solution
Check Mark
To determine

The speed of the ball at the peak of its path.

Answer to Problem 4.56AP

The speed of the ball at the peak of its path is 3gR4 .

Explanation of Solution

Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi , the horizontal range of the ball is R and the maximum height covered by the ball is R6

From part (a) the time of flight is 4R3g .

From the Figure (1) the range of the ball and time of flight is,

R=(vicosθi)t

Rearrange the above equation.

vicosθi=Rt

Substitute 4R3g for t in the above equation.

vicosθi=R4R3g=3gR24R=3gR4

Conclusion:

Therefore, the speed of the ball at the peak of its path is 3gR4 .

(c)

Expert Solution
Check Mark
To determine

The initial vertical component of the velocity.

Answer to Problem 4.56AP

The initial vertical component of the velocity is gR3 .

Explanation of Solution

Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi , the horizontal range of the ball is R and the maximum height covered by the ball is R6

From part (a) vertical component of the initial velocity is gR3 .

Conclusion:

Therefore, the initial vertical component of the velocity is gR3 .

(d)

Expert Solution
Check Mark
To determine

The initial speed of the ball.

Answer to Problem 4.56AP

The initial velocity of the ball is 13gR12 .

Explanation of Solution

Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi , the horizontal range of the ball is R and the maximum height covered by the ball is R6

From part (a) vertical component of the initial velocity is,

visinθi=gR3

Square both side of the above equation.

(visinθi)2=(gR3)2=gR3 (1)

And from part (b) the horizontal component of the velocity is 3gR4

vicosθi=3gR4

Square both side of the above equation.

(vicosθi)2=(3gR4)2=3gR4 (2)

Add equation (1) and (2) to find the initial velocity.

(visinθi)2+(vicosθi)2=3gR4+gR3(vi)2((sinθi)2+(cosθi)2)=9gR+4gR12(vi)2=13gR12vi=13gR12

Conclusion:

Therefore, the initial velocity of the ball is 13gR12 .

(e)

Expert Solution
Check Mark
To determine

The angle θi .

Answer to Problem 4.56AP

The angle θi is 33.7° .

Explanation of Solution

Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi , the horizontal range of the ball is R and the maximum height covered by the ball is R6

From part (a) vertical component of the initial velocity is,

visinθi=gR3

And from part (b) the horizontal component of the velocity is 3gR4

vicosθi=3gR4

Take the ratio of the horizontal component and the vertical component of the initial velocity.

visinθivicosθi=gR33gR4tanθi=4gR9gRθi=tan194=33.7°

Conclusion:

Therefore, the angle θi is 33.7° .

(f)

Expert Solution
Check Mark
To determine

The maximum height that the ball can reach with the same initial velocity.

Answer to Problem 4.56AP

The maximum height that the ball can reach with the same initial velocity is       13R24 .

Explanation of Solution

Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi , the horizontal range of the ball is R and the maximum height covered by the ball is R6

For the maximum height to be gained by the ball the angle made by the horizontal should be 90° .

The formula to calculate the maximum height reached by the projectile is,

Hmax=(vi)2sin290°2g

Here,

Hmax is the maximum height reached by the ball.

Rearrange the above equation.

Hmax=(vi)2sin290°2g=(vi)22g

From part (d) the initial velocity of the ball is,

vi=13gR12

Substitute 13gR12 for vi in the above equation.

Hmax=(13gR12)22g=13gR12(2g)=13R24

Conclusion:

Therefore, the maximum height that the ball can reach with the same initial velocity is       13R24 .

(g)

Expert Solution
Check Mark
To determine

The maximum range of the ball with the same initial velocity.

Answer to Problem 4.56AP

The maximum range of the ball with the same initial velocity is 13R12 .

Explanation of Solution

Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi , the horizontal range of the ball is R and the maximum height covered by the ball is R6

For the maximum range to be gained by the ball the angle made by the horizontal should be 45° .

The formula to calculate the maximum height reached by the projectile is,

Rmax=(vi)2sin22(45°)g=(vi)2g

Here,

Rmax is the maximum range of the ball.

From part (d) the initial velocity of the ball is,

vi=13gR12

Substitute 13gR12 for vi in the above equation.

Rmax=(13gR12)2g=13gR12(g)=13R12

Conclusion:

Therefore, the maximum range of the ball with the same initial velocity is 13R12 .

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Chapter 4 Solutions

PHYSICS 1250 PACKAGE >CI<

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