Question

A parachutist bails out from an aeroplane and after drooping through a distance of $40m$, he opens the parachute and decelerates at $2{m/s}^2$. If he reaches the ground with a speed of $2m/s$ (i) how long he is in air. (ii) at what height did he bail out from the plane?

Answer 1

Applying 3rd equation to calculate the final velocity,

$v^2=0+2\ast 9.8\ast 40$

v = $\sqrt{784}$

= 28 m/s .

Time taken, $t_1=\frac{28}{9.8}$ [ u + at ]

= 2.85 s.

Then, the second part says that he reached the ground with

velocity 2 m$/s$ (final velocity) with deceleration of 2 m$/s^2$

and the initial velocity would be which we calculated above i.e. 28m/s

Assume, it traveled a distance of 'h' while falling.

Again, apply 3rd equation of motion,

$(2{)}^2=(28{)}^2+2(-2)\ast h$

4 * v = 784 - 4

= 780

h = 195m.

Time taken, $t_2=\frac{28-2}{2}$ [ v = u + at ]

= 13 s

Time taken while he is in air,

t = $t_1+t_2$

= 2.85 + 13

= 15.85 s