Answer to Question #100459 in Classical Mechanics for Himanshu

Question

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

Answer 1

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