Answer to Question #100459 in Classical Mechanics for Himanshu

Question #100459
Q2 B) A Man of mass 80 kg dives into a swimming pool from a tower of height 18 m. He was found to go down in water by 2.2 m and then started rising. Find the average resistance of water. Neglect the air resistance of water take value of, g =9.81m/s2
1
Expert's answer
2019-12-16T11:07:28-0500

Formulae adopted is

Final velocity v=√(2gh)

v2-u2=-2as

Resistance=ma

Thus v=√(2x9.81x18)=18.78m/s

This v become u when just touching water surface and then final velocity is zero

Thus

0-18.782=-2a x 2.2

Thus a=80.15m/s2

Thus resistive force=80 x 80.15=6412N



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