# A wooden ball of relative density 0.75 falls into a pond from a height of 1m. if viscous forces due to air and water are neglected, the ball will sink in water to a depth of

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by 2092000

2015-04-12T02:45:04+05:30

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Let the volume of the ball be = V m³
relative density of the wooden ball =  0.75
=>  density d / d_water = 0.75  => density d = 0.75 * d_water

height thru which the ball falls = h1 = 1 m

velocity of the ball when it touches the water surface = v
v² = u² + 2 a s = 0 + 2 g h1
v = √(2 g)

force of buoyancy on the ball = V g * d_water
net force on the ball = buoyancy force - weight  in the upward direction
= V d_water g = V g * 0.75 d_water
=  V g * 0.25  d_water

acceleration of the ball through the water = force / mass
= 0.25 V g d_water / V d  = g /3

apply     v² = u² + 2 a s          =>       0 =  (2 g)  - 2 * g/3  * s
s = 3 meters

Thus the ball sinks to a depth of 3 meters before coming to a stop.  Then it will start slowly moving up due to buoyancy force.

we neglected the drag, and viscous forces in air and in water.

Very nice answee ;)