Question

Two objects having mass in 1 : 4 ratio are dropped
from the same height. The relation between their
velocity when they strike the ground is
(1) Both objects will have the same velocity
(2) The velocity of the first object is twice that of
the second one
(3) The velocity of the 2nd object is one fourth of
that of the 1st object
(4) The velocity of the 2nd object is 4 times that of
the 1st one​

Answer 1

Given :-

Ratio of two objects = 1:4

To Find :-

Relation between velocity

Solution :-

At first the final Velocity

\bf \: v = u + at(1)v=u+at(1)

\bf \: v = u + at(2)v=u+at(2)

Now,

The distance covered

\sf \: s \: = ut + \dfrac{1}{2} {at}^{2} (1)s=ut+21at2(1)

\sf \: s \: = ut + \dfrac{1}{2} {at}^{2} (2)s=ut+21at2(2)

\sf ut \dfrac{a(t_1)}2 =ut \dfrac{a(t_2)}{2}ut2a(t1)=ut2a(t2)

\sf \dfrac{a(t_1)}{2} = \dfrac{a(t_2)}{2}2a(t1)=2a(t2)

Cancelling a and 2

\sf t_1 = t_2t1=t2