Question

A ball of mass m and another ball of mass 2m are dropped from equal height. If times taken by the balls to reach the ground are ${t_1}$ & ${t_2}$ respectively then
(A) ${t_1} = \dfrac{{{t_2}}}{2}$
(B) ${t_1} = {t_2}$
(C) ${t_1} = 4{t_2}$
(D) ${t_1} = \dfrac{{{t_2}}}{4}$

Answer 1

Hint: For calculating the time taken by the body to reach the ground from any height then we use the formula $t = \sqrt {\dfrac{{2h}}{g}} $
When g = gravitational acceleration
h = Height from which particle are dropped.

Complete step by step solution:
Let there are 2 balls of masses m & 2m dropped from height h.



Now for particle of mass m –
$u = 0,$ $t = {t_1}$ & height $ = h,$ $a = g$
So, Acc to newton’s II equation
$s = ut + \dfrac{1}{2}a{t^2}$
$h = 0 + \dfrac{1}{2}gt_1^2$ $ \Rightarrow $ ${t_1} = \sqrt {\dfrac{{2h}}{g}} $ ……(1)
For particle of mass 2m –
$u = 0,$ $t = {t_2},$ height $ = h,$ $a = g$
So, $h = 0 + \dfrac{1}{2}gt_2^2$ $ \Rightarrow $ ${t_2} = \sqrt {\dfrac{{2h}}{g}} $ …..(2)
So, from equation (1) & (2) we can write $\boxed{{t_1} = {t_2}}$

Option B is correct

Note: In free fall motion, if height, form which balls are thrown downward is same, then time taken by them does not depends on masses of particles balls and if heights are different then the ratio of their times are given as $\boxed{\dfrac{{{t_1}}}{{{t_2}}} = \dfrac{{{h_1}}}{{{h_2}}}}$