## College Physics (4th Edition)

(a) The speed of each rock just before it hits the ground is $\sqrt{v_0^2+2gh}$ (b) Using kinematics or conservation of energy both lead to the same result, that the speed of each rock just before it hits the ground is $\sqrt{v_0^2+2gh}$ However, using conservation of energy to solve the question is much easier and much more straightforward.
(a) Let $v_0$ be the initial speed and let $h$ be the height of the cliff. We can use conservation of energy to find the speed of each rock just before it hits the ground: $KE_f = KE_0+U_0$ $\frac{1}{2}mv_f^2 = \frac{1}{2}mv_0^2+mgh$ $v_f^2 = v_0^2+2gh$ $v_f = \sqrt{v_0^2+2gh}$ The speed of each rock just before it hits the ground is $\sqrt{v_0^2+2gh}$ (b) We can use kinematics to solve this question. 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. Using kinematics or conservation of energy both lead to the same result, that the speed of each rock just before it hits the ground is $\sqrt{v_0^2+2gh}$ However, using conservation of energy to solve the question is much easier and much more straightforward.