Question

A man of mass 55 kg climbs up a flight of steps to reach the spring board. The spring board is 6m above the water surface in a swimming pool as shown in the given figure. He jumps up in air, 3m above the spring board, before falling into water in the swimming pool. If the average resisting force exerted by water on the man is 1500N, then the maximum depth of the man in water will be:
a) 2.1 m
b) 3.3 m
c) 4.2 m
d) 5.6 m

Answer 1

The maximum depth of the man in water is 3.3m

Explanation:

Mass of a man,m = $55kg$

He jumps up into the air , 3m above the spring board and the spring board is 6m above the water surface.

Total height of a man from the water surface$= 3+6=9m$

The formula for potential energy,P.E. $= mgh = 55\times10\times9=4950J$

This potential energy would be converted into kinetic energy during fall,

Kinetic energy(K.E.) $= \frac{1}{2}mu^2 = 4950 J$

Velocity at touching water surface, $u^2 = 180 m/s^2$

$= \sqrt{180}\,m/s^2$

Force exerted by the water on the man (Buoyancy of water)

= 1500 N

a is deceleration due to diving of  water

$ma = 1500N$

$m =55kg$

$a= \frac{ -1500}{55} m/s^2$

Final velocity,v = 0

depth of water=s

$v^2 = u^2 - 2as$

$s=-\frac {u^2}{2a} = -\frac{180}{2\times -\frac{1500}{55}} = 3.3\,m$

Thus,the maximum depth of the man in water is 3.3m

Answer 2

Explanation:

When the man jumps ,

total height , h= 6 + 3 = 9 m

initial velocity , u = 0 m/s

acceleration due to gravity, g = 10 m/s^2

Let d be the depth that the man will cover inside the pool

Applying 3rd equation of Motion ,

${v}^{2} - {u}^{2} = 2as$

${v}^{2} = 2(10)(9) = 180 \: m {s}^{ - 1}$

Initial Kinetic Energy of the man as he just reaches the pool

$ke_i = \frac{1}{2}m {v}^{2} = \frac{1}{2} \times 55 \times 180$

$= 4950 \: joules$

Final Kinetic Energy of the man = 0 joules ( as at depth d , its velocity will be 0 )

Work done by the pool on the man = force × displacement

= - 1500d ( force is in upward direction )

By work energy Thereom ,

$work \: done = change \: in \: kinetic \: energy$

$- 1550d = ke_f \: - ke_i$

$- 1550d = 0 - 4950$

$d = \frac{4950}{1500} = 3.3 \: m$

Therefore , it is your answer.

I hope it helps you. If you have any doubts, then don't hesitate to ask.