## College Physics (4th Edition)

The sphere rises to a vertical height of $1.79~m$ above the horizontal surface.
We can use conservation of energy to find the vertical height that the sphere rises: $U_g = KE_{tr}+KE_{rot}$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)~(\frac{v}{R})^2$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$ $Mgh = \frac{7}{10}Mv^2$ $h = \frac{7v^2}{10g}$ $h = \frac{(7)(5.00~m/s)^2}{(10)(9.80~m/s^2)}$ $h = 1.79~m$ The sphere rises to a vertical height of $1.79~m$ above the horizontal surface.