# A parachutist after bailing out falls 50 m without friction. When the parachute opens, it decelerates at 2 m/s². He reaches the ground with a speed of 3 m/s. At what height, did he bail out? (a). 293 m (b). 111 m (c). 91 m (d). 182 m

By BYJU'S Exam Prep

Updated on: September 25th, 2023

It is given that

Distance traveled without friction = 50 m

Acceleration a = – 2 m/s⁻2

Speed before reaching the ground = 3 m/s

We can use the formula

v2 = u2 + 2as

After bailing out from point A, the parachutist falls freely under gravity

Velocity acquired by it will be ‘v’

As u = 0

a = 9/8 m/s2

s = 50 m

Substituting the values

v2 = 0 + 2 x 9.8 x 50

v2 = 980

v = √980 m/s

At point B, the parachute opens and moves with retardation of 2 m/s⁻2 and reaches the ground with a velocity of 3 m/s

Apply the equation for the point BC

v = 3 m/s

u = √980 m/s

a = – 2 m/s⁻2

s = h

Substituting the values

32 = √9802 + 2 x (-2) x h

9 = 980 – 4h

h = (980 – 9)/4

h = 971/4

h = 242.7

h = 243 m (approx)

Total height by which parachutist will bail out = 50 + 243 = 293 m

Therefore, the parachutist will bail out at 293 m.

Summary:

## A parachutist after bailing out falls 50 m without friction. When the parachute opens, it decelerates at 2 m/s2. He reaches the ground at a speed of 3 m/s. At what height, did he bail out? (a). 293 m (b). 111 m (c). 91 m (d). 182 m

A parachutist after bailing out falls 50 m without friction. When the parachute opens, it decelerates at 2 m/s2. He reaches the ground at a speed of 3 m/s. He will bail out at 293 m.

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