Question

A parachutist drops freely from an aeroplane for 10 seconds before he opens the parachute. Then he descents with 2m/s^2 acceleration. His velocity on reaching the ground is 8m/s. The height of the aeroplane is​

Answer 1

Answer:

Distance travelled in first 10 seconds

s=1/2gt^2

s=3.9×100

= 390m

velocity after 10 second

v= u+gt

v= 100m/s

final velocity = 8m/s

distance travelled in next seconds

V^2-u^2=2as

64- 10000= 2×-2×a

s= 2484m.

total height = 2484+500=2984m

Answer 2

Answer:

Distance he falls before the parachute opens is

S₁ = 1/2 x g x 100 = 490m

Then his velocity, u = gt = 98.0m/s

Velocity on reaching = v = 8m/s

Retardation = 2m/s²

v² - u² = 2aS₂

Substitute the above values in the given equation

We get,

⇒ S₂ = 2385m

Therefore,

Total distance (S) = S₁ + S₂

S = 2385 + 490

S = 2875m (height of aeroplane)