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During a tennis match, a player serves the ball at \(23.6 \mathrm{~m} / \mathrm{s}\), with the center of the ball leaving the racquet horizontally \(2.42 \mathrm{~m}\) above the court surface. The net is \(12 \mathrm{~m}\) away and \(0.90 \mathrm{~m}\) high. When the ball reaches the net, (a) does the ball clear it and (b) what is the distance between the center of the ball and the top of the net? Suppose that, instead, the ball is served as before but now it leaves the racquet at \(5.00^{\circ}\) below the horizontal. When the ball reaches the net, (c) does the ball clear it and (d) what now is the distance between the center of the ball and the top of the net?

Short Answer

Expert verified
(a) Yes, the ball clears the net by 0.25 m. (c) No, the ball is 0.46 m below the net.

Step by step solution

01

Calculate Time to Reach Net (Horizontal Serve)

For a horizontal serve, the ball travels at a constant horizontal speed of \( v = 23.6 \mathrm{~m/s} \). The net is \( 12 \mathrm{~m} \) away. We calculate the time \( t \) it takes the ball to reach the net using:\[ d = vt \Rightarrow t = \frac{12 \mathrm{~m}}{23.6 \mathrm{~m/s}} = 0.508 \mathrm{~s} \]
02

Calculate Vertical Position at Net (Horizontal Serve)

The initial vertical position \( y_0 = 2.42 \mathrm{~m} \), and the vertical acceleration \( a = -9.81 \mathrm{~m/s}^2 \) due to gravity. Since the initial vertical velocity \( v_{y0} = 0 \mathrm{~m/s} \), the vertical displacement \( \,\Delta y = v_{y0}t + \frac{1}{2}at^2 \) becomes:\[ \,\Delta y = 0 + \frac{1}{2}(-9.81 \mathrm{~m/s}^2)(0.508 \mathrm{~s})^2 = -1.27 \mathrm{~m} \]Thus, the vertical position at the net is:\[ y = y_0 + \,\Delta y = 2.42 \mathrm{~m} - 1.27 \mathrm{~m} = 1.15 \mathrm{~m} \]
03

Determine Clearance Distance (Horizontal Serve)

The net height is \( 0.90 \mathrm{~m} \). Since the ball's height at the net is \( 1.15 \mathrm{~m} \), it clears the net by:\[ \,\Delta h = 1.15 \mathrm{~m} - 0.90 \mathrm{~m} = 0.25 \mathrm{~m} \]
04

Calculate Initial Vertical and Horizontal Components (Angled Serve)

For the serve angled at \( 5.00^{\circ} \) below the horizontal, we first resolve the initial velocity:\[ v_{x0} = v \cos(5.00^{\circ}) = 23.6 \cos(5.00^{\circ}) \approx 23.53 \mathrm{~m/s} \]\[ v_{y0} = -v \sin(5.00^{\circ}) = -23.6 \sin(5.00^{\circ}) \approx -2.06 \mathrm{~m/s} \]
05

Calculate Time to Reach Net (Angled Serve)

With horizontal velocity \( v_{x0} = 23.53 \mathrm{~m/s} \), the time \( t \) to reach the net remains:\[ t = \frac{12 \mathrm{~m}}{23.53 \mathrm{~m/s}} \approx 0.510 \mathrm{~s} \]
06

Calculate Vertical Position at Net (Angled Serve)

With initial vertical velocity \( v_{y0} = -2.06 \mathrm{~m/s} \), the vertical displacement becomes:\[ \,\Delta y = v_{y0}t + \frac{1}{2}at^2 = (-2.06 \mathrm{~m/s})(0.510 \mathrm{~s}) + \frac{1}{2}(-9.81 \mathrm{~m/s^2})(0.510 \mathrm{~s})^2 \approx -1.98 \mathrm{~m} \]Thus, the vertical position at the net is:\[ y = y_0 + \,\Delta y = 2.42 \mathrm{~m} - 1.98 \mathrm{~m} = 0.44 \mathrm{~m} \]
07

Determine Clearance Distance (Angled Serve)

The net height is \( 0.90 \mathrm{~m} \). Since the ball's height at the net is \( 0.44 \mathrm{~m} \), it does not clear the net and is below the top of the net by:\[ \,\Delta h = 0.90 \mathrm{~m} - 0.44 \mathrm{~m} = 0.46 \mathrm{~m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Velocity
When discussing projectile motion in sports, understanding **horizontal velocity** is crucial. In a tennis serve, the ball leaves the player's racket with a significant horizontal velocity, which is the speed at which the ball travels parallel to the ground. This velocity is constant when air resistance is negligible.
For example, in our tennis problem, the initial horizontal velocity of the ball is given as 23.6 m/s. Since there is no acceleration affecting the ball in the horizontal direction once it leaves the racket, this velocity remains unchanged all the way to the net. You calculate the time it takes for the ball to reach the net by dividing the distance to the net by this horizontal velocity:
  • Distance to net: 12 m
  • Horizontal velocity: 23.6 m/s
  • Time to net: 12 m / 23.6 m/s = 0.508 s
This calculation is fundamental in predicting the ball's motion, as it helps determine when and where the ball will arrive.
Vertical Displacement
**Vertical displacement** refers to how much the ball's height changes from its starting point to where it needs to clear an obstacle, like a net. When a ball is struck horizontally from a height, as in our tennis scenario, gravity causes it to fall as it moves forward.
Initially, when the ball leaves the player's racket, the vertical component of its velocity might be zero if it's served horizontally. This means it only starts to fall due to gravity. However, if it's served at an angle, the vertical component will influence its initial movement.
Using the formula for vertical displacement
  • \(\Delta y = v_{y0}t + \frac{1}{2} a t^2\)
where \(v_{y0}\) is the initial vertical velocity and \(a\) is the acceleration due to gravity The ball's vertical displacement is calculated both when served horizontally and at an angle. This determines how high or low it is at the point it reaches the net.
Gravity
**Gravity** is the universal force that pulls objects toward the center of the Earth. In projectile motion, gravity plays a significant role in determining the trajectory of any object.
It accelerates the ball downward at a constant rate of 9.81 m/s^2, affecting its vertical motion. In the case of our tennis serve, this force doesn't influence the horizontal velocity but instead causes the ball to drop as it travels toward the net.
When calculating projectile motion:
  • This constant downward acceleration makes sure the ball descends from its initial height of 2.42 m.
  • The effect of gravity is more noticeable in the calculations related to the vertical displacement and determines if the ball can potentially drop below the net’s height.
Understanding gravity's impact is essential when predicting whether the tennis ball will clear the net at a given distance.
Trigonometry
**Trigonometry** is a branch of mathematics that deals with angles and lengths. In sports physics, it helps resolve a projectile's velocity into its horizontal and vertical components.
For a serve angled at 5 degrees below the horizontal, trigonometric functions are key:
  • The horizontal component of velocity (\(v_{x0}\)) is found using \(v \cos(\theta)\)
  • The vertical component (\(v_{y0}\)) is calculated with \(-v \sin(\theta)\)
  • These give us the velocities needed to calculate the ball's path
In our tennis scenario, knowing these components allowed us to calculate how the ball would behave differently when served at an angle compared to a straight horizontal serve.
Net Clearance in Tennis
**Net clearance** in tennis is a crucial aspect determining whether a serve is effective and within the rules of the game. It is the distance by which the ball surpasses or falls short of the net height.
In our example:
  • When served horizontally, the ball clears the net by 0.25 m. This means it goes over the net with room to spare, indicating a successful serve.
  • When served at a 5-degree angle downward, the ball fails to clear the net, missing by 0.46 m.
This difference underscores the importance of considering the angle and velocity of a serve to ensure it not only reaches the opponent but stays within play. Assessing net clearance helps fine-tune serves for optimal performance.

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