Aball is thrown vertically upward. after t seconds, its height h (in feet) is given by the function h(in feet) h(t)=100t-16t squared . what is the maximum height that the ball will reach? do not round your answer.

Given the height of the ball at time, t is modeled by:
h(t)=100t-16t^2
the maximum height will be found as follows:
at maximum height:
h'(t)=0
thus given h(t), the derivative will be:
h'(t)=100-32t=0
solving for t we get
t=100/32
thus the time taken for the ball to reach maximum height is t=100/32. Hence the maximum height will be:
h(100/32)=100×100/32-16×(100/32)²
=156.25 ft

It should be a or d i’m not sure