Question

A solid ball of density half that of water falls freely under gravity from a height of 19.6 m and then enters the water. Up to what depth will the ball go? How much time will it take to come again to the water surface. Neglect air resistance and viscosity effects in water. ($g=9.8{ms}^{-2}$)

• A. 4 s
• B. 8 s
• C. 6 s
• D. 2 s

Answer 1

The correct option is A 4 s

Velocity of the ball when hitting water is $v_b=\sqrt{2gh}=\sqrt{2\times 9.8\times 19.6}=19.6$

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, ${\rho }_b=\frac{1}{2}{\rho }_w$

Net force on the ball $f_b=V_b{\rho }_{w}g-V_b{\rho }_{b}g=m_b{a}_b$ where subscript 'b' refers to ball and 'w' refers to

water.

$\therefore V_b{\rho }_b{a}_b=V_b{\rho }_{w}g-\frac{1}{2}{V}_b{\rho }_{w}g=\frac{1}{2}{V}_b{\rho }_{w}g$

$\therefore \frac{1}{2}{V}_b{\rho }_w{a}_b=\frac{1}{2}{V}_b{\rho }_{w}g$

$a_b=g$ upward

since the deceleration is g downward, the ball will go $19.6m$ inside water. Using 1-d kinematics eqn

$s_b=u_{b}t+\frac{1}{2}{a}_b{t}^2$

$19.6=0\times t+\frac{1}{2}{gt}^2$

$\therefore 19.6=\frac{1}{2}\times 9.8\times t^2$

$\Rightarrow t^2=4$

$\Rightarrow t=2$

since the magnitude of acceleration same both downward and upward, the balls has taken 2 secs to reach a depth of $19.6m$. Hence, total time to come back to the surface again after hitting 3 the surface is $2+2=4$ secs