Question

The question is "A man of mass $55kg$climbs upstairs to reach the springboard which is 6m above the surface of a swimming pool. He jumps up into the air, $3m$ above the springboard, before falling into the pool. If the force exerted by water on the man is $1500N$, the maximum depth of the man in water will be"

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