A ball 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
