Word problem/antiderivative

Question

A ball is shot from the ground straight up into the air with initial velocity of 41 ft/sec. Assuming that the air resistance can be ignored, how high does it go?

Hint:The acceleration due to gravity is 32 ft per second squared

Answer 1

Use the energy formula v =sqrt(2gh) where v = velocity, h = height, g = gravitational constant = 32.2 ft/s²

Then, v² = 2gh or h = v2/2g = (41)2/(2)(32.2) = 26.1 ft

Answer 2

s(t) = 41t - 16t²

^{The max occurs when 32t=41 (the derivative = 0)}

s(t) when t = 41/32