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Q.
A ball whose density is $5 \times 10^{3} kg / m ^{3}$ falls into the water from a height of $5 m$. To what depth the ball will go?
Mechanical Properties of Fluids
Solution:
Velocity of ball before entering into the water
$V=\sqrt{2 g h}$
$=\sqrt{2 \times 10 \times 5}=10 m / s$
Due to the upthrust the ball motion will retard.
upthrust $(F)=m a$
Apparent weight $= ma$
$\therefore \left(\frac{\rho-\sigma}{\rho}\right) g = a $
$\Rightarrow \left(\frac{5-1}{5}\right) \times 10=8 m / s ^{2}$
$v^{2}-u^{2}=2 a h$
Due to upthrust after reaching its maximum depth final velocity becomes zero
$(0)^{2}-( V )^{2}=2 as$
$-100=2 \times 8 \times h$
$\frac{-100}{16}=h$
$h=-6.25 m ($ -ve indicates depth $)$