A sphere of radius a and mass m rolls along a horizontal plane with...

From energy conservation
$mgh = \dfrac{1}{2}{mv}_{0}^{2} + \dfrac{1}{2}{I \omega}^{2}$
$⇒ mgh = \dfrac{1}{2}{mv}_{0}^{2} + \dfrac{1}{2} \times \dfrac{2}{5}{ma}^{2} \times {v}_{0}^{2}{a}^{2}$
$⇒ gh = \dfrac{1}{2}{v}_{0}^{2} + \dfrac{1}{5}{v}_{0}^{2}$
$⇒ gh = \dfrac{7}{10}{v}_{0}^{2}$

From triangle, $sin\theta = \dfrac{h}{l}$
Then $h = lsin\theta$
$⇒ lsin\theta = \dfrac{7}{10}{v}_{0}^{2}g$
Thus, the sphere will travel distance, $l = \dfrac{7}{10}\dfrac{{{v}_{0}}^{2}}{gsin\theta}$.