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

Answer:

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

Step-by-step 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 $t=6\ sec$

Answer 2

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.