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(C) Let the velocity of the body just before the collision be $v_1$ and just after the collision be $v_2$.Using the conservation of energy or kinemati

Vedclass · Accepted Answer

$2:1$

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(C) Let the velocity of the body just before the collision be $v_1$ and just after the collision be $v_2$.Using the conservation of energy or kinematic equations:For the fall: $v_1^2 = 2gh_1$,where $h_1 = 10 \ m$.For the rebound: $v_2^2 = 2gh_2$,where $h_2 = 2.5 \ m$.Taking the ratio of the squares of the velocities:$\frac{v_1^2}{v_2^2} = \frac{2gh_1}{2gh_2} = \frac{h_1}{h_2} = \frac{10}{2.5} = 4$.Taking the square root on both sides:$\frac{v_1}{v_2} = \sqrt{4} = 2$.Thus,the ratio of the velocity just before the collision to the velocity just after the collision is $2:1$.

$A$ body falls from a height of $10 \ m$ onto the ground and rebounds to a height of $2.5 \ m$. Find the ratio of the velocity of the body just before the collision and just after the collision.

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