Answer
(a) The ball lands 36.5 meters from the bottom of the cliff.
$v = 28.0~m/s$
(b) When the ball lands on the ground again, the rotational kinetic energy is the same as at the top of the cliff. Therefore, the translational speed is faster when the ball reaches the ground again compared with the initial translational speed. The total energy of the ball is conserved throughout the journey.
Work Step by Step
We can use conservation of energy to find the horizontal speed at the top of the cliff.
$K_2+PE_2 = K_1+PE_1$
$\frac{1}{2}mv_2^2+\frac{1}{2}I\omega_2^2+mgh = \frac{1}{2}mv_1^2+\frac{1}{2}I\omega_1^2+0$
$\frac{1}{2}mv_2^2+\frac{1}{2}(\frac{2}{5}mr^2)(\frac{v_2}{r})^2+mgh = \frac{1}{2}mv_1^2+\frac{1}{2}(\frac{2}{5}mr^2)(\frac{v_1}{r})^2$
$\frac{7}{10}mv_2^2 = \frac{7}{10}mv_1^2-mgh$
$v_2^2 = v_1^2-\frac{10}{7}gh$
$v_2 = \sqrt{v_1^2-\frac{10}{7}gh}$
$v_2 = \sqrt{(25.0~m/s)^2-\frac{10}{7}(9.80~m/s^2)(28.0~m)}$
$v_2 = 15.26~m/s$
We can find the time it takes the ball to fall 28.0 meters.
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}$
$t = \sqrt{\frac{(2)(28.0~m)}{9.80~m/s^2}}$
$t = 2.39~s$
We can find the horizontal distance the ball travels in this time.
$x = v_x~t = (15.26~m/s)(2.39~s)$
$x = 36.5~m$
The ball lands 36.5 meters from the bottom of the cliff.
We can find the vertical speed when the ball hits the ground.
$v_y = g~t = (9.80~m/s^2)(2.39~s)$
$v_y = 23.42~m/s$
We can find the speed of the ball when it hits the ground.
$v = \sqrt{(v_x)^2+(v_y)^2}$
$v = \sqrt{(15.26~m/s)^2+(23.42~m/s)^2}$
$v = 28.0~m/s$
(b) Initially the ball was rotating faster so it had a greater rotational kinetic energy. At the top of the cliff the rotation was slower so the ball had less rotational kinetic energy. When the ball lands on the ground again, the rotational kinetic energy is the same as at the top of the cliff. Therefore, the translational speed is faster when the ball reaches the ground again compared with the initial translational speed. The total energy of the ball is conserved throughout the journey.
