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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