Answer
The sphere rises to a vertical height of $1.79~m$ above the horizontal surface.
Work Step by Step
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.
