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A solid sphere rolls along a horizontal plane with constant speed $v_{0}$ . It encounters an inclined plane at an angle $θ$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will it travel?

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$\frac{10 v_{0}^{2}}{7 g \sin θ}$

$\frac{v_{0}^{2}}{2 g \sin θ}$

$\frac{7 V_{0}^{2}}{10 g \sin θ}$

$\frac{v_{0}^{2}}{5 g \sin θ}$

answer is B.

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Detailed Solution

$S = \frac{V_{0}^{2}}{2 a} = \frac{V_{0}^{2}}{\frac{2. g \sin θ}{1 + \frac{2}{5}}} = \frac{7 V_{0}^{2}}{10 g \sin θ}$

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