Question

A hot air balloon is rising vertically upwards at a constant velocity of $10{\text{ms}}^{-1}$. When it is at a height of $45\text{m}$ from the ground, a man bails out from it. After $3\text{s}$ he opens his parachute and decelerates at a constant rate of $5{\text{ms}}^{-2}$. How far is parachutist from the balloon at $t=3\text{s}$?

• A. $15\text{m}$
• B. $30\text{m}$
• C. $45\text{m}$
• D. $60\text{m}$

Answer 1

The correct option is C $45\text{m}$

Velocity of balloon $=+10\text{m/s}$ (as the velocity of the balloon is upwards)
When the parachutist bails out, he has the velocity of the balloon and has an upward velocity of $10{\text{ms}}^{-1}$, i.e. $u=+10{\text{ms}}^{-1}$.
Also $g=-10{\text{ms}}^{-2}$ (acting downwards).
The displacement of parachutist in $t=3\text{s}$ is given by
$S=ut+\frac{1}{2}g{t}^2$ (By second equation of motion)
$=10\times 3+\frac{1}{2}\times (-10)\times (3{)}^2$
$=-15\text{m}$
Since the displacement is negative, it is directed downwards.
The height by which balloon will rise in time $(t=3\text{s})=10\times 3=30\text{m}$
We know that, the displacement of parachutist in $3\text{s}=-15\text{m}$.
Hence, the parachutist is now $30-(-15)=45\text{m}$ away from the balloon.
Thus, the correct choice is (c).