Question

Question 25
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.

Answer 1

Using the equation of motions to find out the time period of each mass

$S=ut+\frac{1}{2}a{t}^2$

Case 1: $S=h_1$

$h_1=0+\frac{1}{2}g(t_1{)}^2$

$t_1={\left(\frac{2h_1}{g}\right)}^{1/2}$

Case 2: $S=h_2$

$h_2=0+\frac{1}{2}g(t_2{)}^2$

$t_2={\left(\frac{2h_2}{g}\right)}^{1/2}$

Ratio: $\frac{t_1}{t_2}=\frac{(\frac{2h_1}{g}{)}^{1/2}}{(\frac{2h_2}{g}{)}^{1/2}}$

$={\left(\frac{h_1}{h_2}\right)}^{1/2}$

In all the cases, the ratio of time period will be same because time period does not depends on the shape of the body.