Question

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, with size remaining the same in each case? Give reason.

Answer 1

The motion of the object due to free fall:

The distance covered by the object due to the free-fall can be expressed as follows:

$\mathrm{h}=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2$

Here,

$h$ is the distance covered by the object.

$u$ is the initial velocity.

$g$ is the acceleration due to gravity.

$t$ is the time.

Step 1: Calculation of time for object 1.

Since the initial velocity is zero, the time required by the object 1 can be calculated as follows:

$$\begin{gathered} {\mathrm{h}}_1=0\times {\mathrm{t}}_1+\frac{1}{2}{\mathrm{gt}}_1^2 \\ {\mathrm{t}}_1^2=\frac{2{\mathrm{h}}_1}{\mathrm{g}} \end{gathered}$$

Step 2: Calculation of time for object 2.

Since the initial velocity is zero, the time required by the object 1 can be calculated as follows:

$$\begin{gathered} {\mathrm{h}}_2=0\times {\mathrm{t}}_2+\frac{1}{2}{\mathrm{gt}}_2^2 \\ {\mathrm{t}}_2^2=\frac{2{\mathrm{h}}_2}{\mathrm{g}} \end{gathered}$$

Step 3: Calculation of ratio of time required.

$$\begin{gathered} \frac{{\mathrm{t}}_1^2}{{\mathrm{t}}_2^2}=\frac{{\displaystyle \frac{2{\mathrm{h}}_1}{\mathrm{g}}}}{\frac{2{\mathrm{h}}_2}{\mathrm{g}}} \\ \frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}=\frac{\sqrt{{\mathrm{h}}_1}}{\sqrt{{\mathrm{h}}_2}} \end{gathered}$$

Therefore, the ratio of time does not depend upon the mass of the objects, or the shape of the objects.

Hence, for both cases, the ratio of time will remain the same.