Hi there!
Assuming the ball is hollow and spherical:
[tex]I = \frac{2}{3}MR^2[/tex]
Since the ball is both rolling and moving linearly, the total kinetic energy is comprised of both ROTATIONAL and TRANSLATIONAL kinetic energy.
[tex]KE_T = KE_R + KE_{TR}[/tex]
Recall the equations for both:
[tex]KE_{TR} = \frac{1}{2}mv^2\\\\KE_R = \frac{1}{2}I\omega^2[/tex]
We can rewrite the rotational kinetic energy using linear velocity using the following relation:
[tex]\omega = \frac{v}{r}[/tex]
[tex]KE_R = \frac{1}{2}(\frac{2}{3}mR^2(\frac{v^2}{R^2}))\\\\KE_R = \frac{1}{3}mv^2[/tex]
We can now represent the situation as a summation of energies:
[tex].9(KE_R + KE_{TR}) = PE\\\\.9(\frac{1}{2}mv^2 + \frac{1}{3}mv^2) = mgh \\\\.9(\frac{5}{6}mv^2) = mgh[/tex]
Cancel out the mass and rearrange to solve for height:
[tex]\frac{3}{4}v^2 = gh \\\\h = \frac{3v^2}{4g}[/tex]
Plug in the given velocity and g = 9.8 m/s².
[tex]h = \frac{3(8^2)}{4(9.8)} \approx \boxed{4.9 m}[/tex]
