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.

Draw the trajectory of the ball and and indicate the following points: starting point O; the max height A; the terminal point of trajectory B (where the ball hits the ground); C is the position where the ball is after 2.7 sec after the max position (the height of the fence); and point D which is located symmetrically to point C but on OA.

The time of motion from the initial (O) point to the terminal (b) point is t= 2vₒ•sinα/g.
t(OA) =t/2 =2•v₀•sinα/2•g =v(oy)/g =>
v(oy) = g•t (OA)=9.8•3.2 = 31.36 m/s.
t(DA) = t(AC) = 2.7 s. => t(OD) =t(OA)-t(DA) =3.2-2.7= 0.5 s.
The height of the point D= the height of the point C= h
(b) h = v(oy) •t(OD) –gt(OD)²/2 = 31.36•0.5 – 9.8•0.5²/2 = 14.46 m.
(a) vAC: H-h=g•t(AC)²/2 = 35.72 m.
Max height H= h+35.72 = 14.46+32.72 = 50.18 m.

(c)
81 = L/2 + x = v(ox) •31.36/g +v(ox) •2.7
v(ox) •(31.36/9.8 +2.7) = 81
v(ox) = 81•9.8/(31.36 +2.7•9.8) = 13.73 m/s.
X = 13.73•0.5= 6.86 m.

To solve parts (b) and (c) of the problem, we need to use the information provided and apply some principles of projectile motion. Let's break down the problem step by step:

Step 1: Analyze the motion of the ball when it reaches its maximum height.

We know that the ball reaches its maximum height 3.2 seconds after being hit. At the maximum height, the vertical velocity of the ball is zero, and the only force acting on it is the acceleration due to gravity. This means that the time it takes for the ball to reach its maximum height is equal to the time it will take to fall back to its original level.

Since the ball takes 3.2 seconds to reach its maximum height, it will also take 3.2 seconds to fall back down. Thus, the total time of flight for the ball in the air is 3.2 + 3.2 = 6.4 seconds.

Step 2: Calculate the initial vertical velocity and displacement.

Using the equation of motion for vertical displacement, we know that the maximum height reached by the ball can be calculated as:

h = (v0y^2) / (2g)

Where:
h = maximum height reached by the ball
v0y = initial vertical velocity of the ball
g = acceleration due to gravity (9.8 m/s^2)

Since the ball starts at ground level, the initial vertical displacement (h0) is zero. Therefore, we only need to find the initial vertical velocity (v0y).

We know that the time taken to reach maximum height (t) is 3.2 seconds and the acceleration due to gravity (g) is 9.8 m/s^2. Using the equation of motion for vertical velocity:

v0y = gt

Substituting the values, we have:
v0y = (9.8 m/s^2) * (3.2 s)
v0y = 31.36 m/s

Step 3: Calculate the total horizontal displacement and the vertical displacement at the fence.

We are given that 2.7 seconds after reaching its maximum height, the ball barely clears a fence that is 81.0 m from where it was hit. Using the horizontal motion of the ball, we can calculate the total horizontal displacement as:

d = v0x * t

Where:
d = total horizontal displacement
v0x = initial horizontal velocity of the ball
t = total time of flight (6.4 seconds)

The initial horizontal velocity (v0x) remains constant throughout the flight because no horizontal forces act on the ball. Thus, we can determine v0x by considering the horizontal distance covered in the given time:

81 m = v0x * 2.7 s

Solving for v0x:
v0x = 81 m / 2.7 s
v0x = 30 m/s

Lastly, we need to calculate the vertical displacement at the fence. Since the ball reaches its maximum height after 3.2 seconds, it reaches the fence 2.7 seconds after reaching the maximum height. Thus, the vertical displacement at the fence can be calculated as:

h_fence = h0 + v0y * t_fence + (1/2) * g * t_fence^2

Where:
h_fence = vertical displacement at the fence
h0 = initial vertical displacement (0 m)
v0y = initial vertical velocity (31.36 m/s)
t_fence = time taken to reach the fence (2.7 s)
g = acceleration due to gravity (9.8 m/s^2)

Substituting the values, we have:
h_fence = 0 + (31.36 m/s) * (2.7 s) + (1/2) * (9.8 m/s^2) * (2.7 s)^2
h_fence = 84.168 m

Step 4: Final Answers

(a) The maximum height reached by the ball is 50.2 meters.
(b) The height of the fence is 84.168 meters.
(c) The ball strikes the ground 81 meters beyond the fence.