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Standard IX

Physics

First Equation of Motion

An astronaut ...

Question

An astronaut jumps from an airplane. After he has fallen 20 m, his parachute opens. Now he
falls with a retardation of 2 m/s2 and reaches the earth with a velocity of 4.0 m/s. What was the initial height of the airplane? Take „g‟ as 10 m/s2.

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Solution

Before opening the parachute:-
Distance travelled, s= 20 m
u = 0
a = 10 m/s^2
v^2 = u^2+2as
v^2= 0+21020
v^2=400
v=20 m/s

After opening the parachute:-
u=20 m/s
v=4 m/s
a= -2 m/s^2
v^2= u^2+2as
4^2=20^2 - 22s
16=400-4s
4s=400-16=384
s=384/4
s = 96 m

Initial height of the airplane = distance travelled by the astronaut before opening the parachute + distance travelled by the astronaut after opening the parachute.
= 20+96
= 116 m

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An astronaut jumps from an airplane. After he has fallen 20 m, his parachute opens. Now he falls with a retardation of 2 m/s2 and reaches the earth with a velocity of 4.0 m/s. What was the initial height of the airplane? Take „g‟ as 10 m/s2.

An astronaut jumps from an airplane. After he has fallen 20 m, his parachute opens. Now he
falls with a retardation of 2 m/s2 and reaches the earth with a velocity of 4.0 m/s. What was the initial height of the airplane? Take „g‟ as 10 m/s2.