Question

A parachutist after bailing out falls $50\, m$ without friction. When parachute opens, it decelearates at $ 2\,m/s^{2} $ . He reaches the ground with a speed of $3 \,m/s$. At what height, did he bail out?

• A. 91 m
• B. 182 m
• C. 293 m
• D. 111 m

Answer 1

The Correct Option is C

Parachute bails out at height $H$ from ground. Velocity at $A$ $v =\sqrt{2 g h} $ $=\sqrt{2 \times 9.8 \times 50} $ $=\sqrt{980} \,m / s$ The velocity at ground $v_{1}=3\, m / s$ (given) Acceleration $=-2\, m / s ^{2}$ (given)

$ \therefore H-h =\frac{v^{2}-v_{1}^{2}}{2 \times 2} $ $=\frac{980-9}{4} $ $=\frac{971}{4}=242.75$ $\therefore H =242.75+h $ $=242.75+50 \approx 293\, m $