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Question
A healthy youngman standing at a distance of 7 m from a 11.8 m high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms height (1.8 m)?
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Solution
Given:
Height of the building = 11.8 m
Distance of the young man from the building = 7 m
The kid should be caught over 1.8 m from ground.
As the kid is slipping, his initial velocity u is 0.
Acceleration, a = 9.8 m/s2
Let s be the distance before which the kid has to be caught = 11.8 − 1.8 = 10 m
Using the equation of motion, we get:
\[s = ut + \frac{1}{2}a t^2\]
\[\Rightarrow 10 = 0 + \frac{1}{2} \times 9 . 8 \times t^2 \]
\[ \Rightarrow t^2 = \frac{10}{4 . 9} = 2 . 04\]
\[ \Rightarrow t = 1 . 42 s\]
This is the time in which the man should reach the bottom of the building to catch the kid.
Velocity with which the man should run:
\[\frac{s}{t} = \frac{7}{1 . 42} = 4 . 9 \text{ m } /s\]
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