Question

A person standing close to the edge on top of a 112-foot building throws a ball vertically upward. The quadratic function h ( t ) = − 16t^2 + 96t + 112 models the ball's height about the ground, h ( t ) , in feet, t seconds after it was thrown.
a) What is the maximum height of the ball?
b) How many seconds does it take until the ball hits the ground?

Answer 1

Answer:

the maximum height of the ball= 256 feet

7 seconds

Step-by-step explanation:

The quadratic function h ( t ) = − 16t^2 + 96t + 112 models the ball's height about the ground, h ( t ) , in feet, t seconds after it was thrown.

a= -16 , b= 96  and c= 112

To find maximum height , we find vertex

$x= \frac{-b}{2a}=\frac{-96}{2(-16)}= 3$

Now plug in 3 for 't' in h(t)

$h(t) = -16t^2 + 96t + 112$

$h(t) = -16(3)^2 + 96(3)+ 112=256$

Hence vertex is (3, 256)

the maximum height of the ball= 256 feet

(b) when the ball hits the ground then height becomes 0

so we plug in 0 for h(t)  and solve for t

$0 = -16t^2 + 96t + 112$

Apply quadratic formula

$t=\frac{-b+-\sqrt{b^2-4ac}}{2a}$

Plug in all the values a= -16 , b= 96  and c= 112

$t=\frac{-96+-\sqrt{96^2-4(-16)(112)}}{2(-16)}$

t= -1  and t= 7

time cannot be negative. So it will take 7 second to hit the ground.