## College Physics (4th Edition)

(a) The speed of the ball when it hits the ground is $\sqrt{2gh+v^2}$ (b) For all angles $\theta$, the speed of the ball when it hits the ground is $\sqrt{2gh+v^2}$
(a) We can use conservation of energy to find the speed of the ball when it hits the ground: $U_2+KE_2 = U_1+KE_1$ $0+\frac{1}{2}mv_2^2 = mgh+\frac{1}{2}mv^2$ $v_2^2 = 2gh+v^2$ $v_2 = \sqrt{2gh+v^2}$ The speed of the ball when it hits the ground is $\sqrt{2gh+v^2}$ (b) For all angles $\theta$, the speed of the ball when it hits the ground is $\sqrt{2gh+v^2}$. The speed of the ball when it hits the ground does not depend on $\theta$.