A Healthy Young Man 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 At (1.8 M)?

Given:

A healthy young man standing at a distance Of 7m

The height of the building = 11.5m

The kid should be caught over 1.8 m from the ground level

The initial velocity u is 0

Acceleration, a = \( 9.8m/s^{2} \)

The distance before which the kid has been caught be s

\( \Rightarrow 11.8 – 1.8 = 10m \)

Now by using the equation of motion:

\( s = ut + \frac{1}{2}at^{2} \) \( \Rightarrow 10 = 0 + frac{1}{2} * 9.8 * t^{2} \) \( \Rightarrow t^{2} = frac{10}{4.9} = 2.04 \) \( \Rightarrow t = 1.42s \)

The time required for a man to reach the bottom of the building to catch the kid is t = 1.42s.

Now the velocity at which the man should run is/p>

\( \Rightarrow frac{s}{t} \) \( \Rightarrow frac{7}{1.42} \) \( \Rightarrow 4.9m/s^{2} \)

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