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

The potential energy at the top of the track is equal to the kinetic energy at the bottom of the track. Let the height be $h_0$ when the speed at the bottom is $v_0$. $PE_0 = KE_0$ $mg~h_0 = \frac{1}{2}mv_0^2$ $h_0 = \frac{v_0^2}{2g}$ Suppose the speed at the bottom is $2v_0$. We can find an expression for the initial height $h_2$. $PE_2 = KE_2$ $mg~h_2 = \frac{1}{2}m(2v_0)^2$ $h_2 = 4\times~\frac{v_0^2}{2g}$ $h_2 = 4\times h_0$ We need to increase the height of the track by a factor of 4.