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Grade: 9
        
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?
2 years ago

Answers : (1)

Arun
24497 Points
							
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
 
one month ago
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