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 (1.8 m)?

Question

Kevin Nash · Answer

Sol. At point B (i.e. over 1.8 m from ground) the kid should be catched. For kid initial velocity u = 0 Acceleration = 9.8 m/s2 Distance S = 11.8 – 1.8 = 10 m S = ut + ½ at2 ⇒ 10 = 0 + ½ (9.8) t2 ⇒ t2 = 2.04 ⇒ t = 1.42. In this time the man has to reach at the bottom of the building. Velocity s/t = 7/1.42 = 4.9 m/s.

Gayathri Ganesh · Answer

In this case, the height of the man should also be subtracted. If he outstretches his arms to catch the child, the distance between him and the building should be minimised by his arm length. Pls correct me if I'm wrong

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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 (1.8 m)?

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 (1.8 m)?

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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 (1.8 m)?

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 (1.8 m)?

2 Answers

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