91Ó°ÊÓ

1. The depth below the surface from which the ball is released in a pool of water,$\mathrm{Δ²â}=0.600\mathrm{m}$
2. The density of the ball, ${\mathrm{ÒÏ}}_{\mathrm{ball}}=0.300{\mathrm{ÒÏ}}_{\mathrm{w}}$

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, $\mathrm{ÒÏ}=\frac{\mathrm{m}}{\mathrm{V}}$ (ii)

The third equation of motion, ${\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2+2\mathrm{aΔ²â}$ (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,

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

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:

$\begin{array}{c}{\mathrm{v}}_{\mathrm{f}}^2=2\mathrm{aΔ²â}\text{(}âˆ\mu {\text{v}}_{\mathrm{i}}\text{=0)}\\ {\mathrm{v}}_{\mathrm{f}}=\sqrt{2\mathrm{aΔ²â}}\\ =\sqrt{2×22.9\frac{\mathrm{m}}{{\mathrm{s}}^2}×0.600\mathrm{m}}\\ =5.24\mathrm{m}/\mathrm{s}\end{array}$

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

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

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