Answer:
Let m be the mass and p be the density of the ball.
Therefore, density of water = \[2\rho \].
Energy of the ball on striking the water surface = mgh = mg x 19.6
J ...(i)
Net force opposing the motion of the ball in water
\[=uptrust\,-weight\,of\,ball\,=\left( \frac{m}{\rho }\times 2\rho
\times g \right)-mg=mg\]
If d is depth upto which the ball goes in water, then Work done =
energy of the ball on striking water
surface
\[mg\times d=mg\times 19.6\]; \[\therefore
\] \[d=19.6\,m\]
Velocity on striking the water surface \[u=\sqrt{2gh}=\sqrt{2\times
9.8\times 19.6}=19.6m/s\]
From \[\upsilon =u+at\]; \[0=19.6-\left( 9.8
\right)t\] \[\therefore \]t = 2s.
This is the time taken by ball to travel a depth (d) in water.
Total time taken by the ball to reach the water surface = 2 + 2 =
4 s
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