Question

Tanya kicks a ball into the air. The function that models this scenario is f(t) = –16t2 + 96t, where h is the height of the ball in feet and t is the time seconds.

When will the ball hit the ground?

Answer 1

I guess the function is h(t) = -16t^2 + 96t (instead of f(t) )

When the ball hits the groung h(t) = 0

Then solve the equation for t

- 16t^2 + 96 t = 0

Divide by -16

t^2 - 6t = 0

t(t-6) = 0

t =0 is when the ball is kicked off

t = 6 is the answer.

Answer 2

the ball hit the ground at
`t=6\ sec`

Explanation:

we have

`h(t)=-16t^(2) +96t`

This is the equation of a vertical parabola open downward

The vertex is a maximum

we know that

The x-intercept of the function is the value of t when the value of h(t) is equal to zero

The ball hit the ground when h(t) is equal to zero

equate the function to zero and solve for t

`0=-16t^(2) +96t`

Factor the leading coefficient

`-16t(t -6)=0`

The solutions are

`t=0, t=6\ sec`

therefore

the ball hit the ground at
`t=6\ sec`