Question

A balloon rises up with uniform velocity 'u'. A body is dropped from balloon. The time of descent
for the body is

Answer 1

The time of descent for the body is  $h=-ut+\frac{1}{2}gt^{2}$

Given: (i) A balloon which is rising with a uniform velocity 'u'.

          (ii) A body which is dropped from the balloon.

To find: Time of descent for the body.

Explanation:

• Step 1:

Let 'h' be the height from the ground till the balloon rises.                              When the body is dropped from the balloon to the ground it has the                   same velocity (in the upward direction) as the balloon due to law of         inertia that is 'u'.

• Step 2:

The body falling from the balloon will have acceleration due to gravity (in   downward direction) that is 'g'.

Let 't' be the time taken by the body in covering the distance 'h' to the ground.

• Step 3:

Using the equation of motion  $s=ut+\frac{1}{2}at^{2}$

⇒   $-h=ut+\frac{1}{2}( -g)t^{2}$  (Distance 'h' has been taken negative as it had been considered downwards.)

⇒  $h=-ut+\frac{1}{2} gt^{2}$

This is the required time of descent for the body.