Question

The question is "A man of mass 55 kg climbs up stairs to reach the spring board which is 6m above the surface of a swimming pool.He jumps up into the air, 3 m above the spring board, before falling into the pool. If force exerted by water on the man is 1500 N, then maximum depth of the man in water will be" .I need an explained answer

Answer 1

Mass of the man = 55kg
He jumps up into the air , 3m above the spring board and the spring board is 6m above the water surface.
Hence total height = 9m
Velocity of the man when he reaches the surface of the water = √2gs [by v²-u²=2as]

Velocity of the man at the water surface=√2×10×9=√180 m/s²

Force exerted by the water on the man
= 1500 N
Force exerted by the man on the water is mg.
when the man reaches to a depth 'd' its velocity is 0.
That is final velocity is 0 and initial velocity is √180
Let us form a force equation for this,
mg-1500=ma

mg-1500=55[(0-√180)/t]

55(10)-1500=55[(0-√180)/t]

By solving above eq we have,
time taken by man to reach
depth d = t = 0.508 sec
Now,
v²-u² = 2ad

o-(√180)²=2[(0-√180)/0.508]d

d = (90×0.508)/√180

d = 3.409m

Answer 2

Net height of the man from the water surface is 3+6  = 9m.
This potential energy will convert into kinetic energy.
12mu2 = mghwhere v is the velocity 
when swimmer reach to the water surface u= root 2×10×9−−−−−130.41 m/sforce = 1500 
N55a = 1500a = −27.27 m/s2
this is the retardation
v2−u2 = 2as0 −180
= − 1500/55ss =
 180×55/1500s = 6.6 m
where s is the depth in the water.

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