The dimensionless parameters for a ball released and falling from rest in a fluid are

\[ \mathrm{C}_{\mathbf{D}}, \quad \frac{g t^{2}}{D}, \quad \frac{ho}{ho_{b}}, \quad \text { and } \quad \frac{V t}{D} \]

where \(C_{\mathrm{D}}\) is a drag coefficient (assumed to be constant at 0.4), \(g\) is the acceleration of gravity, \(D\) is the ball diameter, \(t\) is the time after it is released, \(ho\) is the density of the fluid in which it is dropped, and \(ho_{b}\) is the density of the ball. Ball 1 , an aluminum ball ( \(ho_{b_{1}}=2710 \mathrm{~kg} / \mathrm{m}^{3}\) ) having a diameter \(D_{1}=1.0 \mathrm{~cm}\), is dropped in water \(\left(ho=1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\). The ball velocity \(V_{1}\) is \(0.733 \mathrm{~m} / \mathrm{s}\) at \(t_{1}=\) \(0.10 \mathrm{~s}\). Find the corresponding velocity \(V_{2}\) and time \(t_{2}\) for ball 2 having \(D_{2}=2.0 \mathrm{~cm}\) and \(ho_{b_{2}}=ho_{b_{1}}\). Next, use the computed value of \(V_{2}\) and the equation of motion,

\[ ho_{b} \forall g-\frac{\mathrm{C}_{\mathbf{D}}}{2} ho A V^{2}=ho_{b} \forall \frac{d V}{d t} \]

where \(V\) is the volume of the ball and \(A\) is its cross-sectional area, to verify the value of \(t_{2}\). Should the two values of \(t_{2}\) agree?