Maximum height is reached when the first derivative is zero and second derivative is -ve.
h(t) = -16t^2 +96t
or h'(t) = -32t +96 = 0
or t = 96/32 = 3 seconds
And h''(t) = -32 < 0, so at time=3 seconds height is maximum.
So, h(3) = 144 feet
(Ans 1) Maximum height reached by the ball is 144 feet
(Ans 2) Time taken to reach maximum height is 3 seconds
f(t) = -5t^2 +20t +5
f'(t) = -10t +20 = 0
or t = 2 seconds
So, f(2) = 25 m
(Ans 3) Time taken to reach maximum height by sam is 2 seconds
(Ans 4) Maximum height reached = 25 m
h(t) = -16t^2 +160t+4
h'(t) = -32t +160 = 0
or t = 5 seconds
So, h(5) = 404 feet
(Ans 5) Time taken by rock to reach maximum height is 5 seconds
(Ans 6) Maximum height attained = 404 feet