A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.2 s after being hit. Then 2.7 s after reaching its maximum height, the ball barely clears a fence that is 81.0 m from where it was hit. Assume the ground is level. (a) What maximum height above ground level is reached by the ball? (b) How high is the fence? (c) How far beyond the fence does the ball strike the ground?

**So far I can only get part (a) which is 50.2 meters.

To solve parts (b) and (c) of the problem, we can use the equations of motion under constant acceleration. The initial velocity, u, is zero since the ball is hit at ground level. The acceleration, a, is due to gravity and is approximately -9.8 m/s^2 (taking it as negative as it opposes the motion).

Let's break down the problem step by step:

Step 1: Determine the time it takes for the ball to reach its maximum height.
Given: Time to reach maximum height, t = 3.2 s

Using the equation:
v = u + at,
where v is the final velocity at maximum height,
u is the initial velocity (which is zero), and
a is the acceleration due to gravity.

We know that the final velocity at maximum height is zero since the ball stops momentarily before falling. So, the equation becomes:
0 = 0 + (-9.8) * t
0 = -9.8t
t = 0 (no real solution)
This means the ball reaches its maximum height instantaneously (at t = 0) and starts to fall immediately.

Step 2: Determine the maximum height, h, reached by the ball.
Given: Time to reach maximum height, t = 3.2 s
Using the equation:
h = ut + (1/2)at^2,
where h is the maximum height,
u is the initial velocity (which is zero), and
a is the acceleration due to gravity.

h = 0 - (1/2) * 9.8 * (3.2)^2
h = 0 - 15.68
h ≈ -15.68 m

Since the height cannot be negative, we discard the negative value to get the maximum height as 15.68 m.

Step 3: Determine the time it takes for the ball to reach the fence.
Given: Time after reaching the maximum height, t = 2.7 s

The total time of flight is the time taken to reach the maximum height, plus the time taken to reach the ground from the maximum height. So, the total time is:
total time = 2 * time to reach maximum height = 2 * 3.2 s = 6.4 s

Since the time measured from the maximum height is 2.7 s, the time remaining to reach the ground is:
remaining time = total time - time at maximum height
remaining time = 6.4 s - 2.7 s
remaining time = 3.7 s

Step 4: Determine the horizontal distance covered by the ball.
Given: Distance from the hitting point to the fence, d = 81.0 m
Time of flight after reaching the maximum height, t = remaining time = 3.7 s

Using the equation:
d = ut + (1/2)at^2,
where d is the horizontal distance covered,
u is the initial velocity (which is zero), and
a is the horizontal acceleration (which is also zero since there is no force acting horizontally).

d = 0 * t + (1/2) * 0 * (t)^2
d = 0

This means that the ball lands exactly at the fence, 81.0 m from where it was hit.

So, to summarize:
(a) The maximum height above ground level reached by the ball is approximately 15.68 m.
(b) The fence is at the same height as the ground.
(c) The ball strikes the ground exactly at the fence, 81.0 m from where it was hit.