First, we need to find the speed of the ball using the equation of motion: \(\sqrt{2gh}\) with g = 9.8 and h=2
\(\sqrt{2*9.8*2}\) = 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 dvg
dvg = ma
a = \(\frac{dvg}{m}\)
Let the density of the ball be d'
m=d'V
a= \(\frac{dvg}{d'v}\)
a= \(\frac{dg}{d'}\)
a= \(\frac{d}{d'}\) g
Relative density d/d' = 0.8
a = \(\frac{g}{0.8}\)
a= \(\frac{9.8}{0.8}\)
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+22.5s
s=8m
The depth to which the ball will sink is 8m
Hope this Helps! :D