# A man of mass 637N dives vertically downwards into a swimming pool from a tower of height 19.6m.

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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.

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Given that:

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

Let, Fr = 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/sec...(ii)

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

Fr – W = ma

Fr – 637 = (637/g) X 96.04

Fr = 6879.6N.

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