Question

A solid ball of density half that of water falls freely under gravity from a height of 19.6 m and then enters water. Neglecting air resistance and viscosity effect in water, the depth up to which the ball will go is (g= 9.8 m//s^(2))

Answer 1

Explanation:

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, $\rho_b= \frac{1}{2}\rho_w </p><p>Net force on the ball f_b= V_b\rho_wg - V_b\rho_bg = m_ba_b$where subscript 'b' refers to ball and 'w' refers to water.

$\therefore V_b\rho_ba_b = V_b\rho_wg - \frac{1}{2}V_b\rho_wg= \frac{1}{2}V_b\rho_wg </p><p> \therefore \frac{1}{2}V_b\rho_wa_b= \frac{1}{2}V_b\rho_wg$

ab=g upward

Since the deceleration is g downward, the ball will go 19.6 m inside water.

Using 1-d kinematics eqn

$s_b= u_bt +\frac{1}{2}a_bt^2 </p><p> 19.6 = 0\times t + \frac{1}{2}gt^2 </p><p> \therefore 19.6 = \frac{1}{2} \times 9.8 \times t^2$

⇒t2=4

⇒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