Suppose that you release a small ball from rest at a depth of $\mathbf{0}\mathbf{.600}\mathbf{}\mathbf{m}$below the surface in a pool of water. If the density of the ball is$\mathbf{0.300}$ that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball.)

1. The depth below the surface from which the ball is released in a pool of water,$\mathrm{\Delta y}=0.600\mathrm{m}$
2. The density of the ball, ${\mathrm{\rho }}_{\mathrm{ball}}=0.300{\mathrm{\rho }}_{\mathrm{w}}$

We can use the concept of Archimedes’ principle and expression of density. When a ball is fully submerged in water, then a buoyant force from the surrounding fluid acts on it. This force has a magnitude equal to the weight of the water displaced by the ball.

Formulae:

Force applied on body (or weight), ${\mathrm{F}}_{\mathrm{b}}={\mathrm{m}}_{\mathrm{f}}\mathrm{g}$ (i)

Density of a substance, $\mathrm{\rho }=\frac{\mathrm{m}}{\mathrm{V}}$ (ii)

The third equation of motion, ${\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2+2\mathrm{a\Delta y}$ (iii)

When the ball is submerged in water, then the buoyant force acts on it. Hence, it accelerates in upward direction. According to Archimedes’ principle,

${\mathrm{F}}_{\mathrm{g}}-{\mathrm{F}}_{\mathrm{b}}={\mathrm{F}}_{\mathrm{net}}$

According to Newton’s second law,

$\begin{array}{c}{\mathrm{m}}_{\mathrm{b}}\mathrm{a}={\mathrm{F}}_{\mathrm{net}}\\ {\mathrm{F}}_{\mathrm{g}}-{\mathrm{F}}_{\mathrm{b}}={\mathrm{m}}_{\mathrm{b}}\mathrm{a}\\ {\mathrm{m}}_{\mathrm{w}}\mathrm{g}-{\mathrm{m}}_{\mathrm{b}}\mathrm{g}={\mathrm{m}}_{\mathrm{b}}\mathrm{a}\\ {\mathrm{\rho }}_{\mathrm{w}}{\mathrm{gV}}_{\mathrm{ball}}-{\mathrm{\rho }}_{\mathrm{b}}{\mathrm{gV}}_{\mathrm{ball}}={\mathrm{\rho }}_{\mathrm{b}}{\mathrm{aV}}_{\mathrm{ball}}\left(\text{Byusingequation(ii)}\right)\\ {\mathrm{\rho }}_{\mathrm{w}}\mathrm{g}-{\mathrm{\rho }}_{\mathrm{b}}\mathrm{g}={\mathrm{\rho }}_{\mathrm{b}}\mathrm{a}\\ \mathrm{a}=\frac{{\mathrm{\rho }}_{\mathrm{w}}\mathrm{g}-{\mathrm{\rho }}_{\mathrm{b}}\mathrm{g}}{{\mathrm{\rho }}_{\mathrm{b}}}\\ =\left(\frac{{\mathrm{\rho }}_{\mathrm{w}}-{\mathrm{\rho }}_{\mathrm{b}}}{{\mathrm{\rho }}_{\mathrm{b}}}\right)\mathrm{g}\\ =\left(\frac{{\mathrm{\rho }}_{\mathrm{w}}}{{\mathrm{\rho }}_{\mathrm{b}}}-1\right)\mathrm{g}\\ =\left(\frac{{\mathrm{\rho }}_{\mathrm{w}}}{0.300{\mathrm{\rho }}_{\mathrm{w}}}-1\right)\mathrm{g}\\ =\left(\frac{1}{0.300}-1\right)9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\\ =22.9\frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

With this acceleration, the ball can be accelerated in upward direction towards the surface of water. Initial velocity of the ball is zero. We can find velocity ${\mathrm{v}}_{\mathrm{f}}$of the ball at the surface of water by using equation (iii) and the given values as:

$\begin{array}{c}{\mathrm{v}}_{\mathrm{f}}^2=2\mathrm{a\Delta y}\text{(}\because {\text{v}}_{\mathrm{i}}\text{=0)}\\ {\mathrm{v}}_{\mathrm{f}}=\sqrt{2\mathrm{a\Delta y}}\\ =\sqrt{2\times 22.9\frac{\mathrm{m}}{{\mathrm{s}}^2}\times 0.600\mathrm{m}}\\ =5.24\mathrm{m}/\mathrm{s}\end{array}$

Now with this velocity ${\mathrm{v}}_{\mathrm{f}}$, the ball goes in upward direction above the surface of water in air. Hence, acceleration due to gravity is acting on it. For this motion of ball, ${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0$ and reaches at maximum height and its final velocity will be zero.

Again using equation (iii) and given values, we get

$\begin{array}{c}0={\mathrm{v}}_0^2-2{\mathrm{gh}}_{\max }\\ {\mathrm{v}}_0^2=2{\mathrm{gh}}_{\max }\\ {\mathrm{h}}_{\max }=\frac{{\mathrm{v}}_0^2}{2\mathrm{g}}\\ =\frac{{(5.24\mathrm{m}/\mathrm{s})}^2}{2\times 9.8\mathrm{m}/{\mathrm{s}}^2}\\ =1.40\mathrm{m}\end{array}$

Hence, the maximum height of the ball is$1.40\mathrm{m}$