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(A) Let the initial horizontal speed of the sphere be $v$.Since the sphere rolls without slipping,its total kinetic energy on the horizontal surface i

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$\sqrt{10 g h / 7}$

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(A) Let the initial horizontal speed of the sphere be $v$.Since the sphere rolls without slipping,its total kinetic energy on the horizontal surface is the sum of its translational and rotational kinetic energies:$KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$For a solid sphere,the moment of inertia $I = \frac{2}{5} m R^2$ and the condition for rolling without slipping is $v = R \omega$,so $\omega = v/R$.Substituting these values:$KE_{total} = \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2 = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2$At the maximum height $h$,the sphere momentarily comes to rest before rolling back down. By the principle of conservation of mechanical energy,the initial kinetic energy equals the potential energy at height $h$:$\frac{7}{10} m v^2 = m g h$Solving for $v$:$v^2 = \frac{10}{7} g h$$v = \sqrt{\frac{10 g h}{7}}$

$A$ solid sphere rolls without slipping,first horizontally and then up to a point $X$ at height $h$ on an inclined plane before rolling down,as shown in the figure below. The initial horizontal speed of the sphere is

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