Answer
$t=2s$
Work Step by Step
Take the derivative of the position function to determine the velocity as a function of time:
$\frac{d}{dt}x(t)=\frac{d}{dt}9.00t-0.750t^{3}$
$v(t)=9.00-2.25t^{2}$
Set the velocity equal to zero, for at the top of its path, the velocity of the ball is 0.
$0=9.00-2.25t^{2}$
Solve for t:
$9.00=2.25t^{2}$
$t^{2}=\frac{9.00}{2.25}$
$t=\sqrt 4$
$t=2s$
