Question

A ball travels on a parabolic path in which the height (in feet) is given by the expression $-25t^2+75t+24$, where $t$ is the time after launch. At what time is the height of the ball at its maximum?

Answer 1

Answer:

The height of the ball is at it's maximum 1.5 units of time after launch.

Step-by-step explanation:

Suppose we have a quadratic function in the following format:

$h(t) = at^{2} + bt + c$

If t is negative, the maximu value of h(t) will happen at the point

$t_{MAX} = -\frac{b}{2a}$

In this question:

$h(t) = -25t^{2} + 75t + 24$

So

$a = -25, b = 75, c = 24$

Then

$t_{MAX} = -\frac{b}{2a} = -\frac{75}{2*(-25)} = 1.5$

The height of the ball is at it's maximum 1.5 units of time after launch.