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 $5{\text{ms}}^{-2}$. After how long does the parachutist takes to hit the ground after his exit from the balloon?

• A. $4\text{s}$
• B. $5\text{s}$
• C. $6\text{s}$
• D. $7\text{s}$

Answer 1

The correct option is B $5\text{s}$

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.
So the height of the parachutist from the ground when he opened his parachute $=45–15=30\text{m}$.
The velocity of the parachutist $3\text{s}$ after he bails out is
$u_2=u+gt$ (Using first equation of motion)
$=10+(-10)\times 3=-20{\text{ms}}^{-1}$ (directed downwards)
At $t=3\text{s}$, the velocity of parachutist is $u_2=-20{\text{ms}}^{-1}$
Let $v_2$ be the velocity with which the parachutist will reach the ground.
Using third equation of motion, we know
$v_2^2-u_2^2=2aS$
$\Rightarrow v_2^2-(-20{)}^2=2(+5\left)\right(-30)$
$\Rightarrow v_2^2=400-300=100$
$\Rightarrow v_2=-10\text{m/s}$ (directed downwards)
Let $t$ be the time taken by the parachutist to reach ground after he opens up his parachute Using first equation of motion, we can say
$v_2=u_2+at$
$\Rightarrow -10=-20+\left(5\right)t$
$\Rightarrow t=2\text{s}$
$\therefore$ The total time the parachutist takes (after his exit from the balloon) to hit the ground is $=3\text{s}+2\text{s}=5\text{s}$
Thus, the correct choice is (b).