An athlete jumping vertically on a trampoline leaves the surface with a velocity of 10 m/s upward. What maximum height does she reach?

Once the athlete leaves the surface of the trampoline, only a conservative force (her weight) acts on her. Therefore, the total mechanical energy of the athlete- Earth system is constant during her flight: ${\mathrm{K}}_{\mathrm{f}} +{\mathrm{U}}_{\mathrm{f}} = {\mathrm{K}}_i+{\mathrm{U}}_{\mathrm{i}}$. Taking the y = 0 at the surface of the trampoline, Ui = mgy = 0. Also, her speed when she reaches maximum height is zero, or ${\mathrm{K}}_{\mathrm{f}}$ = O. This leaves us with ${\mathrm{U}}_{\mathrm{f}} = {\mathrm{K}}_{\mathrm{i}}$ or ${\mathrm{mgy}}_{\max } = \frac{1}{2}{{\mathrm{mv}}_{\mathrm{i}}}^2$, which gives the maximum height as 

${\mathrm{y}}_{\max } =\frac{{{\mathrm{v}}^2}_{\mathrm{i}}}{2\mathrm{g}} = \frac{{(10.0 \mathrm{m}/\mathrm{s})}^2}{2(10.0 \mathrm{m}/{\mathrm{s}}^2)} = 5.0 \mathrm{m}$