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

JEE Main · 2005 · 08 May 2005 · Offline · Kinematics

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates 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✓ Correct
(D)111 m

Answer: (C)

Solution

In the free fall of

50

m the parachutist gains

v_{1} = 2 g \times 50 = 1000

m/s, taking

g = 10

m/s

^{2}

. The parachute then decelerates him at

2

m/s

^{2}

until he lands at

3

m/s, so from

v_{f}^{2} = v_{1}^{2} - 2 a s

we get

9 = 1000 - 4 s

, i.e.

s = \frac{991}{4} \approx 243

m. Adding the two stretches, he bailed out from about

50 + 243 = 293

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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