#### Two objects of masses m1 and m2 having the same size are dropped simultaneously from heights h1 and h2 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.

The weight of an object is nothing but the gravitational force by earth on that body.

The weight is generally given as mg, therefore,

$mg= \frac{GM_{e}m}{R_{e}^{2}}\\ \Rightarrow g=\frac{GM_{e}}{R_{e}^{2}}$

Hence, the acceleration due to gravity does not depend on mass of object.

If we neglect, the friction force of air, mass does not depend on size also.

Whether you drop a solid body or hollow body, their acceleration under gravity will be same in free fall.

Therefore, they will take same time to reach the ground.

The time to fall can be calculated by equation of motion.

$s=ut+\frac{1}{2}at^{2}\\ \Rightarrow h=0t+\frac{1}{2}gt^{2}\\ \Rightarrow t=\sqrt{\frac{2h}{g}}$

For the given objects, time will be:

$\Rightarrow t_{1}=\sqrt{\frac{2h_{1}}{g}}\; \; and\; \; t_{2}=\sqrt{\frac{2h_{2}}{g}}$

$\Rightarrow \frac{t_{1}}{t_{2}}=\sqrt{\frac{h_{1}}{h_2}}\;$

The ratio will not change in either case because acceleration will remain same. In case of free-fall acceleration does not depend upon mass and size of the body