A man of mass 55 kg climbs a flight of steps to r

Question

A man of mass 55 kg climbs a flight of steps to reach the springboard. The springboard is 6 m above the surface of the water. He jumps into air 3 m above the springboard before falling into the swimming pool. The average resisting force exerted on the man by the water is 1500 N the what will be the maximum depth of man in water?

Arun · Accepted Answer

Mass of the man = 55kgHe jumps up into the air , 3m above the spring board and the spring board is 6m above the water surface.Hence total height = 9mVelocity 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/s²Force 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 √180Let us form a force equation for this,mg-1500=mamg-1500=55[(0-√180)/t]55(10)-1500=55[(0-√180)/t]By solving above eq we have,time taken by man to reachdepth d = t = 0.508 secNow,v²-u² = 2ado-(√180)²=2[(0-√180)/0.508]dd = (90×0.508)/√180d = 3.409m

9 grade maths> A man of mass 55 kg climbs a flight of st...

Chelsi Kothari , 7 Years ago

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9 grade maths> A man of mass 55 kg climbs a flight of st...

Chelsi Kothari , 7 Years ago

Arun

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