Question

Questions 1. A parachutists bails out from an aeroplane and after dropping through a distance of 80m opens the parachute and decelerates at 2 m/s sq. If he reaches the ground with a speed of 2m/s, how long is he in the air? At what height did he bail out from the plane.

Answer 1

1st Case (When he jumps out) :-
s = 80m
u = 0m/s
v = ?
a = 10m/s²
s = ut + 1/2at²
80 = 0×t + 1/2×10×t²
80 = 5t²
t² = 80/5
t² = 16
t = 4s
v = u + at
= 0 + 10×4
= 40 m/s
2nd case (when he opens parachute) :-
u = v of 1st case
u = 40 m/s
a = -2 m/s²
v = 2 m/s
v = u + at
2 = 40 + (-2×t)
2 - 40 = -2t
- 38 = -2t
t = 38/2
t = 19 seconds
s = ut + 1/2at²
= 40×19 + 1/2×(-2)×(19)²
= 760 - 361
= 399 m
Height from which he jumped = s1 + s2
= 80 + 399
= 479m

Total time in which he was in air = t1 + t2
= 4 + 19
= 23 seconds