Final answer:
The tennis ball thrown straight up with an initial speed of 22.5 m/s reaches a maximum height of 25.82 meters before coming back down. This is calculated using kinematic equations with an acceleration of -9.81 m/s^2 (gravity) and considering that the final velocity at the peak is 0 m/s.
Step-by-step explanation:
To calculate how high the tennis ball rises when it is thrown straight up with an initial speed of 22.5 m/s, we use the kinematic equations of motion. Since the ball is caught at the same distance above the ground, the final velocity at the peak of its trajectory is 0 m/s. We can use the following kinematic equation:
v^2 = u^2 + 2as
Where v is the final velocity, u is the initial velocity, a is the acceleration (which is -9.81 m/s^2 due to gravity acting downwards), and s is the displacement, which in this case, is the height the ball rises.
We can rearrange the equation to solve for s:
0 = (22.5 m/s)^2 + 2(-9.81 m/s^2)s
Solving for s:
0 = 506.25 m^2/s^2 - 19.62 m/s^2s
s = 506.25 m^2/s^2 / 19.62 m/s^2
s = 25.82 m
Therefore, the tennis ball rises to a maximum height of 25.82 meters above the ground.