Question

Two bodies of different masses ma and mb are dropped from same height h. What is the ratio of the times taken by the two bodies to fall through this distance?

Answer 1

Answer: The ratio of the time is 1:1.

Explanation:

Given that,

Mass of the first body =$M_{a}$

Mass of the second body = $M_{b}$

Height of the both bodies = h

Using second equation of motion

$s = ut+\dfrac{1}{2}at^{2}$

Where, initial velocity u = o

Then,

$s = \dfrac{1}{2}gt^{2}$....(I)

Now, put the value of s in equation (I)

For first body

$h = \dfrac{1}{2}gt_{a}^{2}$

For second body

$h = \dfrac{1}{2}gt_{b}^{2}$

So, the ratio of the time

$\dfrac{h}{h} = \dfrac{t_{a}^{2}}{t_{b}^{2}}$

$\dfrac{t_{a}^{2}}{t_{b}^{2}} = \dfrac{1}{1}$

$\dfrac{t_{a}}{t_{b}} = \dfrac{1}{1}$

Hence, the ratio of the time is 1:1.