Question

Stephen kicked a ball inside. The equation that represents the height h in feet of the ball after being kicked and elapsed amount of time t in seconds

Answer 1

Answer:

The maximum height of the ball is: 14.0625ft

Step-by-step explanation:

The missing information is:

$h(t) = 30t - 16t^2$

Required

The maximum height the ball attained

First, we calculate the time to reach the maximum height.

For a function: $h(t) = at^2 + bt + c$

The maximum is: $t = \frac{-b}{2a}$

So, the maximum of $h(t) = 30t - 16t^2$ is:

$t = \frac{-b}{2a}$

Where

$a = -16; b = 30$

So:

$t = -\frac{30}{2*-16}$

$t = -\frac{30}{-32}$

$t = \frac{30}{32}$

$t = 0.9375$

The maximum height is then calculated as:

$h(t) = 30t - 16t^2$

$h(0.9375) = 30 * 0.9375- 16 * 0.9375^2$

$h(0.9375) = 28.125- 14.0625$

$h(0.9375) = 14.0625$