Answer to Question #91900 in Mechanics | Relativity for harshita

Question

A ball of density "ρ" is dropped from a vertical cliff of height "h" with zero initial velocity. Now it penetrates through water and reach back its surface
Calculate the total time taken. (Note thah "ρ" is less than water density)

Answer 1

We will solve the problem neglecting the measures of the ball and water resistance.

When the ball reaches the surface of the water, it has a velocity $v=\sqrt{2gh}$ and the time needed to reach water $t_1=\sqrt{\frac{2h}{g}}$ . When the ball is in water, then it moves with the acceleration directed upwards

$a=\left(\frac{\rho_w}{\rho}-1\right)g$, where $\rho_w$ is the density of water. The ball stops after the time $t_2=\frac{v}{a}=\sqrt{\frac{2h}{g}}\frac{1}{\frac{\rho_w}{\rho}-1}$ and during this time it passes the distance $s=\frac{v^2}{2a}=\frac{h}{\frac{\rho_w}{\rho}-1}$. To reach back the water surface the ball needs the time $t_3=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2h}{g}}\frac{1}{\frac{\rho_w}{\rho}-1}$ . So the total time taken $t=t_1+t_2+t_3=\sqrt{\frac{2h}{g}}\left(1+\frac{2}{\frac{\rho_w}{\rho}-1}\right)$.