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Chapter 4: Problem 13

A ball rolls off the edge of a horizontal tabletop, \(4.23 \mathrm{ft}\) high. It strikes the floor at a point \(5.11 \mathrm{ft}\) horizontally away from the edge of the table. ( \(a\) ) For how long was the ball in the air? (b) What was its speed at the instant it left the table?

Short Answer

Expert verified
The ball was in the air for approximately \( 0.518 \) seconds and its speed when it left the table was approximately \( 9.886 \mathrm{ft/s} \).

Step by step solution

01

Calculate the Time the Ball was in the Air

To solve this problem, use the equation for vertical motion under constant acceleration due to gravity, \( d = vt + 0.5at^2 \). Since the ball starts from rest, its vertical velocity component \( v = 0 \) at the beginning. The distance \( d \) fallen under gravity is equal to the height of the table, \( 4.23 \mathrm{ft} \). The acceleration \( a \) is due to gravity, and given that gravity's acceleration is \( -32.2 \mathrm{ft/s^2} \), the minus sign indicates that the ball is moving downwards. Setting these values in the equation and solving for \( t \) will give the time that the ball was in the air.
02

Calculation

Setting the values in the equation as described above, \( t = \sqrt{\frac{2 \cdot d}{a}} = \sqrt{\frac{2 \cdot 4.23}{32.2}} = 0.518 \mathrm{s} \). The ball was in the air for 0.518 seconds.
03

Calculate the Initial Speed

We know the horizontal displacement of the ball, \( 5.11 \mathrm{ft} \), and the time it was in the air, \( 0.518 \mathrm{s} \). As the horizontal motion of the ball is uniform, we can use the speed equation \( v = \frac{s}{t}\) to find the initial speed. Here, \( s \) represents the space traveled, and \( t \) is the time.
04

Calculation

Substituting these into the above equation, the initial speed \( v = \frac{5.11}{0.518} = 9.886 \mathrm{ft/s} \) is obtained. Therefore, the initial speed of the ball as it left the table was approximately \( 9.886 \mathrm{ft/s} \).

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