Answer
(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 $\sqrt{2gh+v^2}$
Work Step by Step
(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 $\theta$.
