# Chapter 10 - Energy - Exercises and Problems: 5

(a) The maximum height is 25 meters above the ground. (b) The speed of the ball is 10 m/s (c) The speed at impact is 22 m/s

#### Work Step by Step

(a) Let $KE_1$ be the initial kinetic energy. Let $PE_1$ be the initial potential energy. Let $KE_2$ be the kinetic energy at maximum height. Let $PE_2$ be the potential energy at maximum height. We can use conservation of energy to find the maximum height $h_2$. $KE_2+PE_2 = KE_1+PE_1$ $0+mg~h_2 = \frac{1}{2}mv_1^2+mg~h_1$ $h_2 = \frac{\frac{1}{2}v_1^2+g~h_1}{g}$ $h_2 = \frac{\frac{1}{2}(10~m/s)^2+(9.80~m/s^2)(20~m)}{9.80~m/s^2}$ $h_2 = 25~m$ The maximum height is 25 meters above the ground. (b) Let $KE_3$ be the kinetic energy as the ball is passing the window on the way down. Let $PE_3$ be the potential energy at the height of the window. We can use conservation of energy to find the speed $v_3$. $KE_3+PE_3 = KE_1+PE_1$ Since $PE_3=PE_1$, then $KE_3 = KE_1$. Therefore, $v_3 = v_1$. The speed of the ball is 10 m/s (c) Let $KE_4$ be the kinetic energy at the moment of impact. Let $PE_4$ be the potential energy at ground level. We can use conservation of energy to find the speed at impact $v_4$. $KE_4+PE_4 = KE_1+PE_1$ $\frac{1}{2}mv_4^2+0 = \frac{1}{2}mv_1^2+mg~h_1$ $v_4^2 = v_1^2+2g~h_1$ $v_4 = \sqrt{v_1^2+2g~h_1}$ $v_4 = \sqrt{(10~m/s)^2+(2)(9.80~m/s^2)(20~m)}$ $v_4 = 22~m/s$ The speed at impact is 22 m/s

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