## Physics (10th Edition)

a) $KE=52.2J$ b) $v=48.77m/s$
a) Take the final point to be the highest point the ball reaches. As the ball starts to rise, its KE decreases while its PE increases. From the principle of energy conservation, we have $$KE_f-KE_0 = -(PE_f-PE_0)$$ $$KE_f-\frac{1}{2}mv_0^2=-mg\Delta h$$ $$KE_f=\frac{1}{2}m v_0^2-mg\Delta h$$ We have $g=9.8m/s^2$, $\Delta h=24.6m$, $m=47g=0.047kg$ and $v_0=52m/s$. Therefore, $$KE_f=52.2J$$ b) Similarly, from the principle of energy conservation, we have $$\Delta KE = -\Delta PE$$ $$\frac{1}{2}m\Delta v^2=-mg\Delta h$$ $$\frac{1}{2}\Delta v^2=-g\Delta h$$ We have $g=9.8m/s^2$, $\Delta h=24.6-8=16.6m$ and $\Delta v=v_{f}^2-v_{0}^2=v_{f}^2-52^2=v_{f}^2-2704$. Thus: $$\frac{1}{2}(v_{f}^2-2704)=-162.68$$ $$v_f=\sqrt{-162.68\times2+2704}=48.77m/s$$