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(D) For a block sliding down a smooth inclined plane of height $h$ and angle $	heta$:The acceleration is $a = g \sin 	heta$.The distance traveled al

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Both the blocks will reach the ground with the same speed but the freely falling block will reach first.

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(D) For a block sliding down a smooth inclined plane of height $h$ and angle $	heta$:The acceleration is $a = g \sin 	heta$.The distance traveled along the plane is $s = h / \sin 	heta$.Using $s = \frac{1}{2} a t_s^2$,we get $h / \sin 	heta = \frac{1}{2} (g \sin 	heta) t_s^2$.Thus,$t_s = \sqrt{\frac{2h}{g \sin^2 	heta}} = \frac{1}{\sin 	heta} \sqrt{\frac{2h}{g}}$.The final speed is $v_s = \sqrt{2as} = \sqrt{2(g \sin 	heta)(h / \sin 	heta)} = \sqrt{2gh}$.For a freely falling block:The acceleration is $a = g$.The distance is $h$.Using $h = \frac{1}{2} g t_f^2$,we get $t_f = \sqrt{\frac{2h}{g}}$.The final speed is $v_f = \sqrt{2gh}$.Comparing the times: $t_f / t_s = \sin 	heta$. Since $\sin 	heta Comparing the speeds: $v_f = v_s = \sqrt{2gh}$.Therefore,both reach the ground with the same speed,but the freely falling block reaches first.

$A$ block slides down a smooth inclined plane when released from the top,while another falls freely from the same point. Which of the following statements is correct?

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