Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 2, Problem 63P

(a)

To determine

The distance covered by the ball in first 3s.

(a)

Expert Solution
Check Mark

Answer to Problem 63P

The distance covered by the ball in first 3s is 44.1m.

Explanation of Solution

The acceleration of the ball is 9.80m/s2 and time taken is 3s.

Write the expression to calculate the distance covered.

s=12at2

Here, s is the distance covered, a is the acceleration of the ball and t is the time taken.

Substitute 9.80m/s2 for a and 3s for t in the above equation to calculate s.

s=12(9.80m/s2)(3s)2=44.1m

Conclusion:

Therefore, the distance covered by the ball in first 3s is 44.1m.

(b)

To determine

The speed of the ball.

(b)

Expert Solution
Check Mark

Answer to Problem 63P

The speed of the ball is 7.0m/s.

Explanation of Solution

The distance moved is 2.5m and acceleration of the ball is 9.80m/s2.

Write the expression for the speed of the ball.

v=2ah

Here, v is the speed of the ball and h is the distance moved.

Substitute 2.5m for h and 9.80m/s2 for a in the above equation to calculate v.

v=2(2.5m)(9.80m/s2)=7.0m/s

Conclusion:

Therefore, the speed of the ball is 7.0m/s.

(c)

To determine

The speed of the ball after 3s.

(c)

Expert Solution
Check Mark

Answer to Problem 63P

The speed of the ball after 3s is 29.4m/s.

Explanation of Solution

The initial speed of the ball is 0m/s, the time taken is 3s and acceleration is 9.80m/s2.

Write the expression to calculate the speed of the ball.

v=u+at

Here, v is the speed of the ball, u is the initial speed of the ball and t is the time taken.

Substitute 0m/s for u, 3s for t and 9.80m/s2 for a in the above equation to calculate v.

v=0m/s+9.80m/s2(3s)=29.4m/s

Conclusion:

Thus, the speed of the ball after 3s is 29.4m/s.

(d)

To determine

The position of the ball.

(d)

Expert Solution
Check Mark

Answer to Problem 63P

The position of the ball is 17.1m below the top of the tower.

Explanation of Solution

The initial speed of the ball is 4.80m/s, the time taken is 2.42s and acceleration is 9.80m/s2.

Write the expression for the position of the ball after 2.42s.

y=ut+12at2

Here, y is the position of the ball, u is the initial speed of the ball, t is the time taken.

Substitute 4.80m/s for u, 2.42s for t and 9.80m/s2 for a in the above equation to calculate y.

y=(4.80m/s)2.42s+12(9.80m/s2)(2.42s)2=11.62m28.70m=17.1m

Thus, the ball is 17.1m below the top of the tower.

Conclusion:

Thus, the position of the ball is 17.1m below the top of the tower.

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Chapter 2 Solutions

Physics

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