Question

A ball was kicked into the air from a balcony 20 feet above the ground, and the ball’s height above the ground, in feet, t seconds after the ball was kicked was h(t) = 20 − 16t^2 + 32t. What was the maximum height, in feet, of the ball above the ground after it was kicked?

Answer 1

$h(t)=20-16t^2+32t \\ \Downarrow \\ h(t)=-16t^2+32t+20 \\ \\ a=-16 \\ b=32$

The maximum height is the y-coordinate of the vertex (x,y) of the parabola.
First find the x-coordinate using the formula:
$x=\frac{-b}{2a}=\frac{-32}{2 \times (-16)}=\frac{-32}{-32}=1$

Now plug the value into the equation and find the y-coordinate:
$y=h(x)=h(1)=-16 \times 1^2+32 \times 1+20=-16+32+20=36$

The maximum height is 36 feet.