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

We can use conservation of energy to solve this question. The potential energy at height $h$ will be equal to the kinetic energy at the bottom of the slide. Therefore; $PE = KE$ $mgh = \frac{1}{2}mv^2$ $h = \frac{v^2}{2g}$ $h = \frac{(4.5~m/s)^2}{(2)(9.80~m/s^2)}$ $h = 1.0~m$ The height of the slide should be 1.0 meter.