Question

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

Answer 1

We can start by finding the velocity of the ball while it hits the surface of water. Then we can find the forces acting on the body besides the air resistance and the velocity in water. This helps us to find the net force acting on the ball. We then use the relation between the density of water and the density of the ball to find the acceleration of the ball. We can then apply one of the equations of motion to find the value of time and get the right solution for the question.

Formulas used:
The equation of motion used here is,
$S = ut + \dfrac{1}{2}a{t^2}$
The formula to find the velocity of the ball is,
${v_b} = \sqrt {2gh}$

Complete step by step answer:
We start by mentioning the given data in the question.The initial velocity of the ball is zero as it is falling and not being thrown, $u = 0$.The height from which the ball is thrown is given as, $h = 19.6m$.The net force in the ball is,
${f_b} = {V_b}{\rho _w}g - {V_b}{\rho _b}g = {m_b}{a_b}$
where the subscript “w” and “b” represents the values of water and the ball.

The density of the ball is half that of water,
${V_b}{\rho _b}{a_b} = \dfrac{1}{2}{V_b}{\rho _w}g$
We can find the velocity with which the ball hits the surface of water by using one of the equations of motion ${v^2} - {u^2} = 2aS$.Substituting the values we get,

\Rightarrow {v_b} = \sqrt {2 \times 9.8 \times 19.6} \\\ \Rightarrow {v_b} = 19.6\,m/{s^2}$$ Now that we have the value of velocity, we can move onto finding the acceleration of the ball by using the information given about density. $${V_b}{\rho _b}{a_b} = \dfrac{1}{2}{V_b}{\rho _w}g$$ Cancelling the terms, we get $${a_b} = g$$. Since the deceleration is downward, the ball will go the same distance in the water that is, $$19.6\,m$$Using the equation of motion, $$S = ut + \dfrac{1}{2}a{t^2}$$ The time can be found by, $${t^2} = \dfrac{{2S}}{a} \\\ \Rightarrow t = \sqrt {\dfrac{{2S}}{a}} \\\ \Rightarrow t = \sqrt {\dfrac{{2 \times 19.6}}{{9.8}}} \\\ \therefore t = 2s$$ This is, the ball travels $$19.6\,m$$ downward and upward so the time becomes two times $$2s$$. **Therefore, the total time becomes $$4\,s$$.** **Note:** The forces acting on the ball will be, the buoyancy acting to push the ball upward and the weight of the body that acts downward. These forces are responsible for the net force acting on the ball. The distance travelled by the ball underwater will be the same as $$19.6\,m$$ because the deceleration is downward.