Two bodies $$A$$ (of mass $$1\,kg$$ ) and $$B$$ (of mass $$3\,kg$$ ) are dropped from heights of $$16\,m$$  and $$25\,m,$$  respectively. The ratio of the time taken by them to reach the ground is

Solution :
For free fall from a height, $$u = 0$$  (initial velocity).
From second equation of motion
$$\eqalign{ & h = ut + \frac{1}{2}g{t^2} \cr & {\text{or}}\,\,h = 0 + \frac{1}{2}g{t^2} \cr & \therefore \frac{{{h_1}}}{{{h_2}}} = {\left( {\frac{{{t_1}}}{{{t_2}}}} \right)^2} \cr & {\text{Given,}}\,{h_1} = 16\;m,{h_2} = 25\;m \cr & \therefore \frac{{{t_1}}}{{{t_2}}} = \sqrt {\frac{{{h_1}}}{{{h_2}}}} = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5} \cr} $$
NOTE
Time taken by the object in falling does not depend on mass of object.