physics help please?

Question

A golf ball is hit at ground level. The ball is
observed to reach its maximum height above
ground level 4.3 s after being hit. 0.61 s after
reaching this maximum height, the ball is
observed to barely clear a fence that is 186 ft
from where it was hit.
The acceleration of gravity is 32 ft/s^2.
How high is the fence? Answer in units of
ft.

Answer 1

First find the maximum height. You have final velocity (0), acceleration (32) and time (4.3), which is all you need to find height. use the equation:

height = (final vel) * time + .5 * acceleration * time^2

For the second part, initial velocity is 0, acceleration is again 32, and the time is .61.

height = (start height) - (initial vel) * time - .5 * acceleration * time^2

Answer 2

This can be solved by only looking at the dynamics of the vertical, or y:

y(t)=v*t-1/2*g*t^2
where
v is the vertical component of velocity and gi is gravity
v(t)=v-*g*t
at t=4.3, v=0
v=32*4.3
=137.6 ft/sec

At t= 4.3+.61
=4.91
y(4.91)=137.6*4.91-16*4.91^2
=290ft

j

Answer 3

The answer is 0.5 x32 ( t1^2 -t2^2) = 16 ( 4.3^2 - 0.61^2) = 16*4.91*3.69 = 289.9 ft.

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Maximum height reached is H = 0.5 g t*t. Where t = 4.3s

H = 295.84 ft.

The time taken to reach the top of the fence is 0.61s.

During this time the ball falls vertically through a distance S.

S = 0.5 g t*t.
S = 5.95 ft.

Height of the fence = 295.84 - 5.95 = 289.9 ft.

Answer 4

The Beatles!

Did I guess right? What did I win???