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(B) When the ball is moving upward,both gravity and the drag force act downwards. The equation of motion is: $m \frac{dv}{dt} = -mg - m\gamma v^2$. Di

Vedclass · Accepted Answer

$\frac{1}{\sqrt{\gamma g}} 	an^{-1} \left( \sqrt{\frac{\gamma}{g}} V_0 ight)$

Vedclass · Answer

(B) When the ball is moving upward,both gravity and the drag force act downwards. The equation of motion is: $m \frac{dv}{dt} = -mg - m\gamma v^2$. Dividing by $m$,we get: $\frac{dv}{dt} = -(g + \gamma v^2)$. Rearranging for integration: $dt = -\frac{dv}{g + \gamma v^2} = -\frac{1}{\gamma} \frac{dv}{(g/\gamma) + v^2}$. Integrating from $t=0$ to $T$ and $v=V_0$ to $0$: $\int_0^T dt = -\frac{1}{\gamma} \int_{V_0}^0 \frac{dv}{(g/\gamma) + v^2}$. Let $a^2 = g/\gamma$,then $\int \frac{dv}{a^2 + v^2} = \frac{1}{a} 	an^{-1}(\frac{v}{a})$. $T = \frac{1}{\gamma} \int_0^{V_0} \frac{dv}{(g/\gamma) + v^2} = \frac{1}{\gamma} [\frac{1}{\sqrt{g/\gamma}} 	an^{-1}(\frac{v}{\sqrt{g/\gamma}})]_0^{V_0}$. $T = \frac{1}{\gamma} \sqrt{\frac{\gamma}{g}} 	an^{-1} (V_0 \sqrt{\frac{\gamma}{g}}) = \frac{1}{\sqrt{\gamma g}} 	an^{-1} (V_0 \sqrt{\frac{\gamma}{g}})$.

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