Problem 36 - Fundamentals of Physics

Question

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 \(2.37 \mathrm{~m}\) 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?

Step-by-Step Solution

Answer

(a) Yes, the ball clears the net. (b) The distance is 0.17 m. (c) No, it does not. (d) The distance is 0.867 m.

Step 1: Determine time to reach the net

First, calculate how long it takes for the ball to travel 12 meters horizontally to reach the net. Since the horizontal speed is constant, \( v_x = 23.6 \, \text{m/s} \), use the formula for horizontal motion: \( t = \frac{d}{v_x} \).\[ t = \frac{12}{23.6} \approx 0.5085 \, \text{s} \]

Step 2: Calculate vertical displacement

Using the time calculated in Step 1 and considering the initial vertical velocity is zero as the ball leaves horizontally, use the formula for vertical motion: \( y = y_0 + v_{0y}t - \frac{1}{2}gt^2 \). With \( y_0 = 2.37 \, \text{m}, \ v_{0y} = 0, \text{ and}\ g = 9.8 \, \text{m/s}^2 \), the vertical displacement is: \[ y = 2.37 - \frac{1}{2} \times 9.8 \times (0.5085)^2 \approx 1.07 \, \text{m} \]

Step 3: Check if the ball clears the net

Compare the vertical position of the ball when it reaches the net with the height of the net: \( 1.07 \, \text{m} \) versus \( 0.90 \, \text{m} \). Since \( 1.07 > 0.90 \), the ball clears the net.

Step 4: Compute distance between ball and net

The distance between the center of the ball and the top of the net is the difference in height: \[ 1.07 \, \text{m} - 0.90 \, \text{m} = 0.17 \, \text{m} \]

Step 5: Calculate initial vertical velocity for angled serve

For the angled serve, the initial velocity components are \( v_x = 23.6 \cos(-5^\circ) \) and \( v_{0y} = 23.6 \sin(-5^\circ) \). \[ v_{0y} = 23.6 \times -0.0872 \approx -2.057 \, \text{m/s} \]

Step 6: Determine vertical displacement for angled serve

Using the formula for vertical motion again with the initial vertical velocity from Step 5, find the new vertical position: \( y = y_0 + v_{0y}t - \frac{1}{2}gt^2 \). \[ y = 2.37 - 2.057 \times 0.5085 - \frac{1}{2} \times 9.8 \times (0.5085)^2 \approx 0.033 \, \text{m} \]

Step 7: Verify if ball clears net with angled serve

Check the ball's height at the net for the angled serve: \( 0.033 \, \text{m} \) versus \( 0.90 \, \text{m} \). Since \( 0.033 < 0.90 \), the ball does not clear the net.

Step 8: Find distance to top of net for angled serve

The distance between the center of the ball and the top of the net is the difference in height: \[ 0.90 \, \text{m} - 0.033 \, \text{m} = 0.867 \, \text{m} \]

Key Concepts

Trajectory

Trajectory is a core concept in projectile motion. It refers to the path that an object follows when it moves under the influence of gravity alone, after being projected into the air. The object in our scenario, a tennis ball, travels in a curved trajectory known typically as a parabolic path.
For a horizontally served ball, like in our exercise, the trajectory is influenced by its initial horizontal velocity. When the ball is served at an angle, the trajectory becomes more complex, influenced by both horizontal and vertical components of velocity. This path determines whether the ball will clear the net or strike it, which is a critical aspect of the given physics problem.
The shape of the trajectory is dictated by several factors including initial velocity, the angle of projection, and gravitational acceleration. Understanding this parabolic curve helps in predicting the motion of any projectile accurately.

Initial Velocity

The initial velocity of a projectile is the speed at which it is launched. It is divided into horizontal and vertical components. In our scenario, for the horizontal serve:

• Initial horizontal velocity remains constant because there is no horizontal force acting on the ball other than air resistance, which is negligible in a basic physics problem like this.
• Initial vertical velocity is zero, as the serve is horizontally launched.

In altering conditions where the ball is served at a 5-degree angle:

• The initial horizontal velocity decreases slightly due to the angle, calculated using cosine: \(v_x = 23.6 \, \text{m/s} \times \cos(-5^\circ)\).
• Initial vertical velocity is not zero anymore but instead a small negative value. It is calculated with sine: \(v_{0y} = 23.6 \, \text{m/s} \times \sin(-5^\circ)\).

These components influence how far and high the projectile, in this case, the tennis ball, will travel before gravity brings it back to the ground. Understanding the breakdown of initial velocity into its components is crucial for solving physics problems involving projectile motion.

Vertical Displacement

Vertical displacement in projectile motion refers to the change in vertical position of the projectile from its initial launch position to any given point in time.
In our problem, we investigate how far the tennis ball drops due to gravity by the time it reaches the net. This vertical drop is calculated by:

• Using the formula \( y = y_0 + v_{0y}t - \frac{1}{2}gt^2 \), where \(y_0\) is initial height, \(v_{0y}\) initial vertical velocity, and \(g\) acceleration due to gravity.

For the horizontal serve, vertical displacement is governed solely by gravity since the initial vertical velocity is zero.
For the 5-degree angled serve, initial downward vertical velocity adds additional displacement to that caused by gravity. This makes the ball reach a lower point by the time it arrives at the net.
Knowing how to calculate vertical displacement helps solve many problems related to whether or not a projectile will clear an object like the net in a tennis serve.

Physics Problem Solving

Successfully solving physics problems involves understanding and applying several fundamental concepts. For our exercise regarding the tennis serve, the following steps are essential:

• Identifying the known quantities such as initial speed, heights, and distances.
• Breaking down the initial velocity into horizontal and vertical components if the projectile is angled.
• Calculating time of flight using the horizontal motion equation \(t = \frac{d}{v_x}\).
• Determining vertical displacement using gravitational acceleration and motion equations.

Applying these steps systematically ensures all aspects of projectile motion, from trajectory to vertical displacement, are addressed.
Moreover, iterative evaluation, like checking whether a ball clears an obstacle, is often required in such problems. This reinforces the importance of careful calculations and understanding of the problem context to achieve the correct solution.
Each problem is unique, thus necessitating a problem-solving approach tailored to the specific details and requirements of the scenario.