physics problem max height?

Question

A golf ball is hit at ground level. The ball is
observed to reach its maximum height above
ground level 8:6 s after being hit. 1.2 s after
reaching this maximum height, the ball is
observed to barely clear a fence that is 413 ft
from where it was hit.
The acceleration of gravity is 32 ft/s2 :
How high is the fence? Answer in units of
ft.

Answer 1

Ignoring air resistance - which, of course, is a major factor in the flight path of golf balls - we assume that the vertical velocity is only affected by gravity. Let the initial vertical velocity be u.

Now, since the ball reaches vertical velocity 0 at time 8.6 seconds, we have u + 8.6(-32) = 0 => u = 275.2 ft/s (!). At time 9.8 seconds the vertical displacement (i.e. height) of the ball will be
s = ut + at^2 / 2 = 275.2 (9.8) + (-32) (9.8^2) / 2
= 1160 ft.

That's one high fence. But then, the ball was hit impressively hard...

Answer 2

413ft x 32ft/s2 = 422,912 / 1.2s 352,426.666........ /413 = 853.33 feet per sec2 max hight of 853.33 feet @ 32 ft per sec @ a distyances of 413ft

Max hight 1,706.66 ft
Wall hight 1160 ft
I think????? I'm missing something here Check this twice befor you use it

Answer 3

You can't solve that w/o the maximum height.