Good luck.
-Andy
The ball goes up, reaches its maximum height, stops, then starts coming down again.
The operative word there is "stops".
So if you can find the value of h when t is equal to zero you're made.
From the equations of motion you've got
h = u t + 1/2 a t^2
where u = 320, a = -32 and h is the value you're trying to find. This matches the equation you've been given, except for the 5. So write it:
h - 5 = -16 t^2 + 320 t
The 5 signifies the extra 5 feet above the ground you start throwing the ball upwards from.
So now you can look at the equation of motion that goes:
v^2 = u^2 + 2 a (h-5)
You've got v, it's equal to zero (see above), and h is what you want to find.
So you have h-5 = (v^2 - u^2) / (2 a)
and plugging in the numbers you get
h-5 = (0 - 320^2) / (2 x (-32))
that should be enough, you should be able to work it out from there.