Two objects of masses $m_1$ and $m_2$ having the same size are dropped simultaneously from heights $h_1$ and $h_2$ respectively. Find out the ratio of time they would take in reaching the ground. Will this ratio remain the same if
$(i)$ one of the objects is hollow and the other one is solid and
$(ii)$ both of them are hollow, size remaining the same in each case. Give reason.

As given, two objects of masses $m_1$ and $m_2$ having the same size are dropped simultaneously from heights $h_1$ and $h_2$ respectively. 

Here for mass $m_1$:

Initial velocity $u=0$

Acceleration $a=g$

Distance $s=h_1$

Let the time taken is $t_1$

On using second equation of motion, $h=ut+\frac{1}{2}gt^2$

Or $h_1=0+\frac{1}{2}gt_1^2$

Or $h_1=\frac{1}{2}gt_1^2$

Or $t_1=\sqrt{\frac{2h_1}{g}}$

Similarly, for an object having mass $m_2$, let $t_2$ is taken to reach the ground.

So, $t_2=\sqrt{\frac{2h_2}{g}}$

Now, $\frac{t_1}{t_2}=\frac{\sqrt{\frac{2h_1}{g}}}{\sqrt{\frac{2h_2}{g}}}$

Or $\frac{t_1}{t_2}=\sqrt{\frac{h_1}{h_2}}$

So, the ratio of time will remain the same if $(i)$ one of the objects is hollow and the other one is solid and $(ii)$ both of them are hollow, the size remaining the same in each case.

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Updated on: 10-Oct-2022

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