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 is

A ball of mass m is dropped from a cliff of height H. The ratio of its

1.	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 is
Answer» `3:4` `4:3` `1:3` `3:1` Solution :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`