Question

Q. A ball of relative density 0.8 falls into water from a height of 2m. find the depth to which the ball will sink ?

Answer 1

First, we need to find the speed of the ball using the equation of motion:
`√(2gh)` with g = 9.8 and h=2


`√(2*9.8*2)` = 6.32 m/s

Now we need to calculate the force which the water used to try to stop the ball using the Buoyancy force d*v*g
d*v*g = ma
a =
`(dvg)/(m)`

Let the density of the ball be d'
m=d'V
a=
`(dvg)/(d'v)`
a=
`(dg)/(d')`
a=
`(d)/(d')` * g

Relative density d/d' = 0.8
a =
`(g)/(0.8)`
a=
`(9.8)/(0.8)`

a = 12.25
`m^(2)`/s
deceleration a'=a-g = 2.45
`m^(2)`/s

The final speed of the ball is v'=0
-v'^2 = v^2 + 2a's
40 = 0+2*2.5s
s=8m


The depth to which the ball will sink is 8m

Hope this Helps! :D