## College Physics (4th Edition)

The ball's kinetic energy when it hits the ground will be equal in magnitude to the sum of the initial gravitational potential energy and the part of the elastic potential energy that was transferred to the ball. We can find the speed of the ball when it hits the ground: $\frac{1}{2}mv^2 = mgh+(\frac{1}{2}kx_1^2-\frac{1}{2}kx_2^2)$ $mv^2 = 2mgh+[k~(x_1^2-x_2^2)]$ $v^2 = \frac{2mgh+[k~(x_1^2-x_2^2)]}{m}$ $v = \sqrt{\frac{2mgh+[k~(x_1^2-x_2^2)]}{m}}$ $v = \sqrt{\frac{(2)(0.056~kg)(9.80~m/s^2)(1.4~m)+[(28~N/m)~((0.18~m)^2-(0.12~m)^2)]}{0.056~kg}}$ $v = 6.04~m/s$ The speed of the ball when it hits the ground is 6.04 m/s.