Question

A ball is thrown upward from ground level. Its height h, in feet, above the ground after t seconds is h-48t -16t^2.

Find the maximum height of the ball.

Answer 1

36 feet.

Explanation:

We have been given that a ball is thrown upward from ground level. Its height h, in feet, above the ground after t seconds is
`h(t)=-48t -16t^2`. We are asked to find the maximum height of the ball.

We can see that our given equation is a downward opening parabola, so its maximum height will be the vertex of the parabola.

To find the maximum height of the ball, we need to find y-coordinate of vertex of parabola.

Let us find x-coordinate of parabola using formula
`x=-(b)/(2a)`.

`x=-(-48)/(2(-16))`

`x=-(48)/(32)`

`x=-(3)/(2)`

So, the x-coordinate of the parabola is
`-(3)/(2)`. Now, we will substitute
`x=-(3)/(2)` in our given equation to find y-coordinate of parabola.

`h(t)=-48t -16t^2`

`h(-(3)/(2))=-48(-(3)/(2))-16(-(3)/(2))^2`

`h(-(3)/(2))=-24(-3)-16((9)/(4))`

`h(-(3)/(2))=72-4*9`

`h(-(3)/(2))=72-36`

`h(-(3)/(2))=36`

Therefore, the maximum height of the ball is 36 feet.