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(B) Since the hill is smooth, the potential energy at the top is converted into kinetic energy at the point where the height is $\frac{H}{2}$.Let $v$

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$H$

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(B) Since the hill is smooth, the potential energy at the top is converted into kinetic energy at the point where the height is $\frac{H}{2}$.Let $v$ be the velocity at height $\frac{H}{2}$. By the law of conservation of energy:$mgH = mg(\frac{H}{2}) + \frac{1}{2}mv^2$$mg(\frac{H}{2}) = \frac{1}{2}mv^2$$v = \sqrt{gH}$Now, the object acts as a horizontal projectile from a height $h' = \frac{H}{2}$.The time taken to reach the ground is given by:$t = \sqrt{\frac{2h'}{g}} = \sqrt{\frac{2(H/2)}{g}} = \sqrt{\frac{H}{g}}$The horizontal distance (range) covered is:$x = v 	imes t = \sqrt{gH} 	imes \sqrt{\frac{H}{g}} = H$

$A$ small object slides down with initial velocity equal to zero from the top of a smooth hill of height $H$. The other end of the hill is horizontal and is at height $\frac{H}{2}$ as shown in the figure. The horizontal distance covered by the object from the end of the hill to the ground is

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