Question

A balloon is released from the ground $(h=0\text{m})$ from rest and starts accelerating with acceleration $5{\text{m/s}}^2$ vertically upwards. An object is dropped from the balloon, when it is at height $10\text{m}$ from the ground. If the balloon is released at $t=0\text{sec}$, then find the time when the object reaches the ground. Take $g=10{\text{m/s}}^2$.

• A. $2.73\text{s}$
• B. $4.73\text{s}$
• C. $4\text{s}$
• D. $2\text{s}$

Answer 1

The correct option is B $4.73\text{s}$

When object is dropped from balloon, $h=10\text{m}$
and let velocity of balloon be $v\text{m/s}$
Given $a=+5{\text{m/s}}^2$
Taking vertically upwards as $+ve$ y-direction and applying kinematic equation:
$v^2=u^2+2ah$
$v^2=0+2\times 5\times 10$
$\therefore v^2=100\Rightarrow v=10\text{m/s}$

Using 2nd eqn of motion
$h=u{t}_1+\frac{1}{2}a{t}_1^2$
$+10=0+\frac{1}{2}\times 5\times t_1^2$
$\Rightarrow t_1^2=4$
or $t_1=2\text{s}$

For released object, initial velocity will be equal to that of velocity of balloon
$\Rightarrow u=10\text{m/s}$
Let it reach the ground after time $t_2\text{s}$.
Then, displacement of object $S=-10\text{m}$
Applying 2nd eqn of motion
$S=u{t}_2+\frac{1}{2}a{t}_2^2$
$-10=+10t_2-\frac{1}{2}\times 10\times t_2^2[\because a=-g=-10{\text{m/s}}^2]$
$\Rightarrow 5t_2^2-10t_2-10=0$
$\therefore t_2=1+\sqrt{3}=2.73\text{s}$

Hence, total time $t=t_1+t_2=2+2.73=4.73\text{s}$