Person standing on top of a 56 ft building throws a ball vertically upward. The quadratic function h=-16t^2+104t+56 models the ball's height about the ground, h in feet, t seconds after it was thrown

Question

What is the maximum height of the ball?

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

Answer 1

take the derivative of h(t) and set it equal to zero to find time when it reaches the maximum height

h'(t) = -32t + 104 =0

t=104/32 =13/4 = 3.25 seconds to reach max height. Plug that into the h(t) equation

h(3.25) =-16(13/4)^2 + 104(13/4) + 56 = 225 feet = max height

it hits the ground in 7 seconds

set h(t) = 0 and solve for t. use the quadratic formula

-16t^2 + 104t +56 = 0

t = -104/-32 + (1/32)sqr(104^2+4(64)(56)) = (104+120)/32 = 7 seconds to hit the ground