find the maximum height

Question

the function s(t) = -16t^2 + v₀t + s₀ where s(t) is the height of the object in feet, v₀ is the initial velocity, s₀ is the initial height and t is the time in seconds

A ball is shot upwards from the surface of the earth (s₀=0) with an initial velocity of 45 feet per second. What is the maximum height reached by the ball?

round to the nearest integer

Answer 1

s(t) = -16t² + 45t, velocity v(t) = s'(t) = -32t + 45 = 0 at top point; from here t = 45/32 = 1.41 s

Then maxinum height h_max = s(1.41) = -16·1.41² + 45·1.41 = 31.64 feet

Answer 2

s(t) = -16t² + 45t + 0 given data

Maximum occurs at t = -b / 2a, or t = -45/-32