A solid ball of density half that of water falls freely under gravity from a height of $19.6 m$ and then enter water. The time (in seconds) it will take to come again to the water surface is (Neglect air resistance & velocity effects in water)

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Solution

Velocity of the ball when hitting water is $v_{b} = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 19.6} = 19.6$
when the ball enters water, the following forces will act on the ball.
a) weight of the ball downward
b) buoyancy force on the ball upward
Given, $ρ_{b} = \frac{1}{2} ρ_{w}$
Net force on the ball $f_{b} = V_{b} ρ_{w} g - V_{b} ρ_{b} g = m_{b} a_{b}$ where subscript 'b' refers to ball and 'w' refers to water.
$∴ V_{b} ρ_{b} a_{b} = V_{b} ρ_{w} g - \frac{1}{2} V_{b} ρ_{w} g = \frac{1}{2} V_{b} ρ_{w} g$
$∴ \frac{1}{2} V_{b} ρ_{w} a_{b} = \frac{1}{2} V_{b} ρ_{w} g$
$a_{b} = g$ upward
Since the deceleration is g downward, the ball will go 19.6 m inside water.
Using 1-d kinematics eqn
$s_{b} = u_{b} t + \frac{1}{2} a_{b} t^{2}$
$19.6 = 0 \times t + \frac{1}{2} g t^{2}$
$∴ 19.6 = \frac{1}{2} \times 9.8 \times t^{2}$
$\Rightarrow t^{2} = 4$
$\Rightarrow t = 2$
Since the magnitude of acceleration same both downward and upward, the balls has taken 2 secs to reach a depth of 19.6 m.
Hence, total time to come back to the surface again after hitting 3 the surface is $2 + 2 = 4$ secs

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A solid ball of density half that of water falls freely under gravity from a height of 19.6m and then enter water. The time (in seconds) it will take to come again to the water surface is (Neglect air resistance & velocity effects in water)

A solid ball of density half that of water falls freely under gravity from a height of $19.6 m$ and then enter water. The time (in seconds) it will take to come again to the water surface is (Neglect air resistance & velocity effects in water)