Question

An aeroplane drops a parachutist. After covering a distance of 40 m, he opens the parachute and retards at 2 ms^-2. If he reaches the ground with a speed of 2 m^-1, he remains in the air for about

• A. 16s
• B. 3s
• C. 13 s
• D. 10 s

Answer 1

Answer: (A)

16s

Answer 2

Using h = 1/2 gt², we get, t₁ = $\sqrt{2h/g}$.

Let t₁ be the time taken from instants of jumping to the opening of parachute, then

t₁ = $\sqrt{\frac{2 \times 40}{9.8}}$ = 2.86 s

His velocity at this point is given by,

v₁² = 2gh₁ = 2 × 9.8 × 40

784 or v₁ = 28 ms^-1

For the remaining journey ,

v = v₁ + at₂

or t₂ = $\frac{v - v_{1}}{a} = \frac{2 - 28}{- 2} = 13s$

∴ total time = t₁ + t₂ = 2.86 + 13

= 15.86 = 16s