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A ball of mass m is dropped from a cliff of height H. The ratio of its kinetic energy to the potential energy when it is fallen through a height 3/4 H isA. `3:4`B. `4:3`C. `1:3`D. `3:1` |
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Answer» Correct Answer - C Total mechanical energy at height, H `E_H=mgH` Let `v_h` be velocity of the ball at height h `(=3/4H)` `therefore` Total mechanical energy at height h, `E_h=mgh+1/2mv_h^2` According to law of conservation of mechanical energy, `E_H=E_h, mgH = mgh+1/2mv_h^2` `v_h^2=2g(H-h)` Required ratio of kinetic energy to potential energy at height h is `K_h/V_h=(1/2mv^2h)/(mgh) =(1/2m2g(H-h))/(mgh)=(H/h-1)=1/3` |
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