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Chapter 4: Problem 25

(a) During a tennis match, a player serves at \(23.6 \mathrm{~m} / \mathrm{s}\) (as recorded by radar gun), with the ball leaving the racquet \(2.37 \mathrm{~m}\) above the court surface, horizontally. By how much does the ball clear the net, which is \(12 \mathrm{~m}\) away and \(0.90 \mathrm{~m}\) high? (b) Suppose the player serves the ball as before except that the ball leaves the racquet at \(5.0^{\circ}\) below the horizontal. Does the ball clear the net now?

Short Answer

Expert verified
In part (a), the ball clears the net. However, in part (b), the ball does not clear the net.

Step by step solution

01

Find Time to Reach the Net in Part (a)

The time it takes for the ball to reach the net can be calculated using the formula for uniform motion: \(d = vt\), where \(d\) is distance (12 m), \(v\) is velocity (23.6 m/s), and \(t\) is time. Solve this equation for \(t\): \(t = d/v\).
02

Calculate Descent of the Ball in Part (a)

Now we find out how much the ball would have fallen during this time. We use the equation of motion: \(d = ut +(1/2)gt^2\), where \(u\) is initial vertical velocity (0), \(g\) is acceleration due to gravity (9.8 m/s²), \(t\) is time from Step 1, and \(d\) is the distance (descent). As the initial vertical velocity (\(u\)) of the ball is 0, this leaves us with the equation \(d =(1/2)gt^2\). From this, we find the total descent of the ball.
03

Clearance of the Ball Over the Net in Part (a)

Now, to find out whether the ball clears the net or not, we subtract the height of the net (0.90 m) from the initial height of the ball (2.37 m) and the descent calculated in the previous step. If the result is positive, the ball clears the net, else it does not.
04

Find Time to Reach the Net in Part (b)

Using the same method as in Step 1, we can find the time it takes for the ball to reach the net if served at an angle. We use the same formula but the horizontal velocity (v) now becomes the horizontal component of the total velocity which is \(23.6 \cos (5.0) \) m/s. So, \(t = d/v\).
05

Calculate Descent of the Ball in Part (b)

In this case, the initial vertical velocity (\(u\)) of the ball is \(23.6 \sin(5.0)\) m/s (downward). Using the case (b) time found from Step 4 in the earlier equation of motion, we can find the total descent of the ball.
06

Clearance of the Ball Over the Net in Part (b)

We use the same way as in Step 3 to find out whether the ball clears the net or not. We subtract the height of the net (0.90 m) from the initial height of the ball (2.37 m) and the descent calculated in the previous step. If the result is positive, the ball clears the net, else it does not.

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