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

Answer: (C)

293 m

Answer 2

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