Question

A man of mass 637N dives vertically downwards into a swimming pool from a tower of height 19.6m. He was found to go down in water by 2m and then started rising. Find the average resistance of the water. Neglect the resistance of air.

Answer 1

Given that: 

W = 637N, h = 19.6m, S = 2m, g = 9.8m/sec² 

Let, F_r = Average resistance 

Initial velocity of man u = 0,

 V² = u² + 2gh 

= 0 + 2 X 9.8 X 19.6 

V = 19.6 m/sec ...(i) 

Now distance traveled in water = 2m,v = 0, u = 19.6m/sec now apply 

V² = u² – 2as 

0 = 19.62 – 2a X 2

a = 96.04m/sec² ...(ii)

Since, net force acting on the man in the upward direction = F_r – W But the net force acting on the man must be equal to the product of mass and retardation.

F_r – W = ma

F_r – 637 = (637/g) X 96.04

F_r = 6879.6N.