## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The kinetic energy at the bottom of the quarter pipe should be equal to the potential energy at the top of the quarter pipe which is at the height $h = 3.0~m$. Therefore; $KE = PE$ $\frac{1}{2}mv^2 = mgh$ $v^2 = 2gh$ $v = \sqrt{2gh}$ $v = \sqrt{(2)(9.80~m/s^2)(3.0~m)}$ $v = 7.7~m/s$ The skateboarder needs a speed of 7.7 m/s at the bottom.