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.

Answer 1

Answer:

36 feet.

Step-by-step 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 $x=-\frac{b}{2a}$.

$x=-\frac{-48}{2(-16)}$

$x=-\frac{48}{32}$

$x=-\frac{3}{2}$

So, the x-coordinate of the parabola is $-\frac{3}{2}$. Now, we will substitute $x=-\frac{3}{2}$ in our given equation to find y-coordinate of parabola.

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

$h(-\frac{3}{2})=-48(-\frac{3}{2})-16(-\frac{3}{2})^2$

$h(-\frac{3}{2})=-24(-3)-16(\frac{9}{4})$

$h(-\frac{3}{2})=72-4*9$

$h(-\frac{3}{2})=72-36$

$h(-\frac{3}{2})=36$

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