Question

When a balloon rising vertically upwards at a velocity of 10 $m{s}^{-1}$ is at height of 45 m from the ground, a parachutist bails out from the balloon after 3 s, he opens the parachute and decelerates at a constant rate of 5 $m{s}^{-2}$.
(a) What is the height of parachutist above the ground when he opens the parachute?
(b) How far is he from the balloon at this instant?
(c) With what velocity does he strike the ground?
(d) What time does he take in striking the ground after his exit from the balloon? (g= 10 $m{s}^2$)

Answer 1

Ans. (a) 30 m, (b) 45 m, (c) 10 $m{s}^{-1}$, (d) 5 s
Velocity at F= 3 sec,
$mg{h}_1+\frac{1}{2}m{V}_1^2=mg{h}_2+\frac{1}{2}m{V}_2^2\left(10\right)\left(45\right)+\frac{1}{2}\left(100\right)=10\left(30\right)+\frac{1}{2}{V}^{2}V=20m/snow,a=-5m/ss=30m\Rightarrow V^2-u^2=2as\Rightarrow V^2-400=-2\left(5\right)\left(30\right)V^2=100\Rightarrow V=10M/s$
(d) v=u+at
$\Rightarrow$ 10=20-5t
t=2 sec
but initially he took 3 sec to on parachute
$\Rightarrow$ t=2+3=5 sec