## College Physics (4th Edition)

(a) In each of the three cases, the final speed is $\sqrt{v_0^2+2gh}$. Thus all three situations result in the same final speed. (b) The final speed for each rock is $19.8~m/s$
(a) Let down be the positive vertical direction. Let $h$ be the height of the cliff. Let $v_0$ be the rock's initial velocity. Note that $a_y = g$. We can find the final vertical velocity of the rock that is thrown straight down: $v_{y,f}^2 = v_0^2+2gh$ $v_{y,f} = \sqrt{v_0^2+2gh}$ Since the horizontal component of velocity is zero, the speed just before hitting the ground is $\sqrt{v_0^2+2gh}$ We can find the final vertical velocity of the rock that is thrown straight up: $v_{y,f}^2 = (-v_0)^2+2gh$ $v_{y,f} = \sqrt{(-v_0)^2+2gh}$ $v_{y,f} = \sqrt{v_0^2+2gh}$ Since the horizontal component of velocity is zero, the speed just before hitting the ground is $\sqrt{v_0^2+2gh}$ We can find the final vertical velocity of the rock that is thrown horizontally: $v_{y,f}^2 = v_0^2+2gh$ $v_{y,f} = \sqrt{0+2gh}$ $v_{y,f} = \sqrt{2gh}$ We can find the final speed of the rock: $v_f = \sqrt{v_0^2+v_{y,f}^2}$ $v_f = \sqrt{v_0^2+(\sqrt{2gh})^2}$ $v_f = \sqrt{v_0^2+2gh}$ In each of the three cases, the final speed is $\sqrt{v_0^2+2gh}$. Thus all three situations result in the same final speed. (b) We can find the final speed of each of the three rocks: $v_f = \sqrt{v_0^2+2gh}$ $v_f = \sqrt{(10.0~m/s)^2+(2)(9.80~m/s^2)(15.0~m)}$ $v_f = 19.8~m/s$ The final speed for each rock is $19.8~m/s$