A tennis player stands 15.0m from the net. They hit a ball from a height of 2.50m above the ground so that initially it has a horizontal velocity directed parallel to the ground toward the net. What speed does the ball have to have to just clear the net (which is 0.9m high)? How long is the ball in the air and where does the ball hit the ground relative to the service line 7.0m from the far side of the net?

Question

A tennis player stands 15.0m from the net. They hit a ball from a height of 2.50m above the ground so that initially it has a horizontal velocity directed parallel to the ground toward the net. What speed does the ball have to have to just clear the net (which is 0.9m high)? How long is the ball in the air and where does the ball hit the ground relative to the service line 7.0m from the far side of the net?

Answer 1

To find the speed the ball needs to clear the net, we can use the concept of projectile motion. In this case, the ball follows a parabolic trajectory. Here's how you can solve this problem step by step:

1. Start by finding the time it takes for the ball to reach the net's height. We can use the equation for vertical displacement:
h = ut + (1/2)gt^2

Here, h is the vertical displacement (0.9m), u is the initial vertical velocity (which is 0 since the ball starts from rest), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.

Rearrange the equation to solve for t:
t = sqrt(2h/g)

Plug in the values:
t = sqrt(2 * 0.9 / 9.8) = sqrt(0.1836) ≈ 0.428 s

2. Now, find the horizontal distance the ball travels during this time. We can use the equation for horizontal displacement:
s = vt

Here, s is the horizontal distance (15m), v is the horizontal velocity (which is what we need to find), and t is the time we found in step 1.

Rearrange the equation to solve for v:
v = s / t

Plug in the values:
v = 15 / 0.428 ≈ 35.05 m/s

So, the speed the ball needs to have to just clear the net is approximately 35.05 m/s.

3. Next, let's find the total time the ball is in the air. Since the ball follows a parabolic trajectory, the time it takes to hit the ground is twice the time it takes to reach the net's height. So, the total time in the air is:
2 * t = 2 * 0.428 ≈ 0.856 s

The ball is in the air for approximately 0.856 seconds.

4. Finally, let's determine where the ball hits the ground relative to the service line.
The horizontal distance covered by the ball in the air can be found using the equation:
s = ut + (1/2)at^2

Since the initial horizontal velocity is directly towards the net (parallel to the ground), there is no acceleration in the horizontal direction, so a = 0. Therefore, the equation becomes:
s = ut

Plug in the values:
s = v * t
s = 35.05 * 0.856 ≈ 30.02 m

So, the ball hits the ground approximately 30.02m away from the server line, towards the far side of the net.