A tennis ball is thrown up with an initial speed of 22.5m/s. It is caught at the same distance above the ground.

A) how high does the ball rise
B) how long does the ball remain in the air?

To answer both parts of the question, we can use the equations of motion under constant acceleration. The acceleration in this case would be due to gravity, and we can assume that air resistance is negligible.

A) To find the maximum height reached by the tennis ball, we can use the formula for displacement:

s = ut + (1/2)at^2

where:
s = displacement or height
u = initial velocity
t = time
a = acceleration

In this case, the initial velocity (u) is 22.5 m/s and the acceleration (a) is -9.8 m/s^2 (negative sign because gravity acts downward). We want to find the height (s) when the velocity (v) is zero, so the time at the maximum height can be found using the formula:

v = u + at

0 = 22.5 - 9.8t

Solving this equation for t gives:

t = 22.5/9.8 ≈ 2.30 s

Now we can substitute this time value back into the equation for displacement:

s = ut + (1/2)at^2

s = 22.5(2.30) + (1/2)(-9.8)(2.30)^2

s ≈ 26.00 m

Therefore, the maximum height reached by the ball is approximately 26.00 meters.

B) The total time the ball remains in the air is the time it takes to reach the maximum height (upward) plus the time it takes to return to the ground (downward). As we have already calculated the time it takes to reach the maximum height (approximately 2.30 s), we need to find the time it takes to return to the ground.

To find this time, we can use the equation:

s = ut + (1/2)at^2

where s = 0 (because the ball is back on the ground), u = 0 (because the ball starts and ends at rest), and a = -9.8 m/s^2. Solving for t gives:

0 = (1/2)(-9.8)t^2

0 = -4.9t^2

t^2 = 0

t = 0

We can see that the time taken to return to the ground is 0 seconds. Therefore, the total time the ball remains in the air is approximately 2.30 seconds.

To summarize:
A) The ball rises to a height of approximately 26.00 meters.
B) The ball remains in the air for approximately 2.30 seconds.

initial KE=final PE

1/2 m 22.5^2=m g h
solve for h.

time in air:
hf=hi+vi*t-4.9t^2
hf=hi and vi is given, solve for t.