Two objects of masses m_1 and m having the same size are dropped … - Vidyadip

Question

Two objects of masses $m_1$ and $m$ 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

One of the objects is hollow and the other one is solid.
Both of them are hollow, size remaining the same in each case. Give reason.

✓

Answer

From secound equation of motion$, \text{h}=\text{ut}+\frac{1}{2}\ \text{gt}^2$
so, $\text{h}=\frac{1}{2}\ \text{gt}^2$
Where$, h =$ Displacement
$u =$ initial velocity
$t =$ Time and $g =$ Acceleration due to gravity
$\Rightarrow\text{gt}^2=2\text{h}$
$\Rightarrow\text{t}^2=\frac{2\text{h}}{\text{g}}$
$\text{t}=\sqrt{\frac{2\text{h}}{\text{g}}}$
$\therefore$ Time taken by first boject of mass$, m_1$
$\text{t}_1=\sqrt{\frac{2\text{h}_1}{\text{g}}}\ [$for $II$ object $u = 0]$
Similarily time taken by object of mass $m_2\text{t}_2=\sqrt{\frac{2\text{h}_2}{\text{g}}}$
$\Rightarrow\frac{\text{t}_1}{\text{t}_2}=\sqrt{\frac{\text{h}_1}{\text{h}_2}}$
Acceleration due to gravity is independent of mass of falling body.
so, the ratio remains the same. if bodies are hollow, then also ratio remains the same
i.e., $\text{t}_1:\text{t}_2=\sqrt{\text{h}_1}:\sqrt{\text{h}_2}$

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