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Chapter 3: Problem 42

You serve a tennis ball from a height of \(1.8 \mathrm{~m}\) above the ground. The ball leaves your racket with a speed of \(18.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(7.00^{\circ}\) above the horizontal. The horizontal distance from the court's baseline to the net is \(11.83 \mathrm{~m},\) and the net is \(1.07 \mathrm{~m}\) high. Neglect spin imparted on the ball as well as air resistance effects. Does the ball clear the net? If yes, by how much? If not, by how much did it miss?

Short Answer

Expert verified
Answer: To determine if the tennis ball clears the net and by how much, follow the steps provided in the solution. Calculate the initial horizontal and vertical velocities, the time it takes to reach the net horizontally, and the vertical position when it reaches the net. Compare the calculated vertical position to the height of the net to find if it clears the net and by how much.

Step by step solution

01

Identify the given information

We are given the following information: - Initial height of the ball: \(1.8 \mathrm{~m}\) - Initial speed of the ball: \(18.0 \mathrm{~m} / \mathrm{s}\) - Launch angle: \(7.00^{\circ}\) - Distance to the net: \(11.83 \mathrm{~m}\) - Height of the net: \(1.07 \mathrm{~m}\)
02

Calculate the initial horizontal and vertical velocities

Using the launch angle and the initial speed, we can find the initial horizontal and vertical components of velocity using the trigonometric functions sine and cosine: - Initial horizontal velocity (\(v_{0x}\)) = \(18.0 \mathrm{~m/s} \times \cos (7.00^{\circ})\) - Initial vertical velocity (\(v_{0y}\)) = \(18.0 \mathrm{~m/s} \times \sin (7.00^{\circ})\)
03

Calculate the time it takes to reach the net horizontally

We can find the time it takes for the ball to travel horizontally to the net by dividing the distance to the net by the initial horizontal velocity. Let's represent this time as \(t\). \(t = \frac{11.83 \mathrm{~m}}{v_{0x}}\)
04

Calculate the vertical position when it reaches the net

Using the time, \(t\), we can find the vertical position, \(y\), when the ball reaches the net. For this, we'll use the equation \(y = y_0 + v_{0y}t - \frac{1}{2}gt^2\), where \(y_0\) is the initial height, \(v_{0y}\) is the initial vertical velocity, and \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)). \(y = 1.8 \mathrm{~m} + v_{0y}t - \frac{1}{2}(9.81 \mathrm{~m/s^2})t^2\)
05

Compare the calculated vertical position to the height of the net

Now that we have the vertical position of the ball as it reaches the net, we can compare it to the height of the net to determine if it clears the net and by how much. - If \(y > 1.07 \mathrm{~m},\) the ball clears the net and the difference is given by \(y - 1.07 \mathrm{~m}\) - If \(y < 1.07 \mathrm{~m},\) the ball doesn't clear the net and the difference is given by \(1.07 \mathrm{~m} - y\) Using the given information and following these steps, we can determine if the ball clears the net and by how much.

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Most popular questions from this chapter

Two swimmers with a soft spot for physics engage in a peculiar race that models a famous optics experiment: the Michelson-Morley experiment. The race takes place in a river \(50.0 \mathrm{~m}\) wide that is flowing at a steady rate of \(3.00 \mathrm{~m} / \mathrm{s} .\) Both swimmers start at the same point on one bank and swim at the same speed of \(5.00 \mathrm{~m} / \mathrm{s}\) with respect to the stream. One of the swimmers swims directly across the river to the closest point on the opposite bank and then turns around and swims back to the starting point. The other swimmer swims along the river bank, first upstream a distance exactly equal to the width of the river and then downstream back to the starting point. Who gets back to the starting point first?

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