Aperson 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?

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

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

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

Apply quadratic formula

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

t= -1  and t= 7

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

the answer is 489

step-by-step explanation:

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