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
[tex]x= \frac{-b}{2a}=\frac{-96}{2(-16)}= 3[/tex]
Now plug in 3 for 't' in h(t)
[tex]h(t) = -16t^2 + 96t + 112[/tex]
[tex]h(t) = -16(3)^2 + 96(3)+ 112=256[/tex]
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
[tex]0 = -16t^2 + 96t + 112[/tex]
Apply quadratic formula
[tex]t=\frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]
Plug in all the values a= -16 , b= 96 and c= 112
[tex]t=\frac{-96+-\sqrt{96^2-4(-16)(112)}}{2(-16)}[/tex]
t= -1 and t= 7
time cannot be negative. So it will take 7 second to hit the ground.
