Question: A soldier jumps out from an aeroplane with a parachute. After dropping through a distance of 19.6 m,...

Question

A soldier jumps out from an aeroplane with a parachute. After dropping through a distance of 19.6 m, he opens the parachute and decelerates at the rate of 1 ms^-2. If he reaches the ground with a speed of 4.6 ms^-1, how long was he in air?

Answer 1

Solution

Let t₁ be the time before opening of parachute.

Using h = ut + 1/2 gt², we get

19.6 = 0 + 1/2 × 9.8 × t₁²

or t₁² = 19.6/4.9 = 4 or t₁ = 2 s

Taking v₁ as velocity attained after falling through 19.6 m and using v² – u² = 2gh, we have

v₁² - 0 = 2 × 9.8 × 19.6

Again taking t₂ as time taken after opening of parachute and using v = u + at, we get

4.6 = 19.6 – 1 × t₂

or t₂ = 19.6 – 4.6n = 15 s

∴ total time = t₁ + t₂ = 2 + 15 = 17s.