Math problem speed and height

Question

A person standing close to the edge on top of a 108-foot building throws a ball vertically upward. The quadratic function  models the ball's height above the ground, , in feet,  seconds after it was thrown.a)

What is the maximum height of the ball?​  

 feet

b) How many seconds does it take until the ball hits the ground? 

  seconds

Answer 1

h(t) = -16t^2 + 132t + 108 = 0 (for when it hits the ground, solve for t)

divide by 4

-4t^2 + 33t + 27 = 0

use the quadratic formula to solve for t if you can't factor, but this factors fairly easily

(4t+3)(-t+9) = 0

set each factor = 0 and solve for t

t=9 and -3/4 seconds, ignore the negative solution

It hits the ground in 9 seconds

max height is found by taking the derivative and setting it =0

-32t +132 = 0

t=132/32 = 33/8 = 4.125 seconds

plug 4 1/8 into the original equation and solve for max height

-16(4.125)^2 +132(4.125) + 108 = max height =380.25 feet