A person standing close to the edge on top of a 48-foot building throws a ball vertically upward.

Question

 The quadratic function

h(t) = 16t+52t + 48 models the ball's height about the ground, h(t) in feet, t seconds after it was thrown. 

What is the maximum height of the ball?  (Write the answer to the nearest hundredth) 

Answer 1

h = -16t^2 +52t + 48

take the derivative and set =0

h' =-32t +52 = 0

t =52/32 = 13/8 seconds to reach max height

max h = -16(13/8)^2 + 52(13/8) + 48

= -169/4 + 84.5 + 48

= -42.25 + 132.5

= 90.25 feet = maximum height

that's not just the nearest hundredth, its the exact answer

(aside from using an approximation for the effect of gravity)