Answer
$t_{up} = V_{y0}$$/g ; t_{down} = V_{y0}$$/g$
Work Step by Step
Choose x = 0 and y = 0 to be the point of launch and upward to be the positive y direction. Let the initial vertical velocity be $V_{y0}$. Use equation 2.11a to find the time to reach the highest point, at which the vertical velocity is zero. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. The time moving up is $V_{y0}$$/g$.
Now use equation 2.11c to find the maximum height reached: this is $\frac{Vy0^{2}}{2g}$.
Finally, use equation 2.11b to find the time for the object to fall that distance with a starting vertical velocity of 0. The time to fall is $V_{y0}$$/g$, which was to be proven.
