Laws of Motion Part 1

Calculation: Given, Initial velocity of the ball, $u = V_{0}$ Drag force, $F_{d}=m y v^{2}$ Velocity at a maximum height $=0$ Where $m$ is the Mass of the ball, $v$ is its instantaneous velocity and $y$ is a constant We know that acceleration is the change in velocity. Force is given by, $F = ma$ Thus, the net force on the ball: $F_{\text{net}} = mg + myv^{2}$ $F_{\;\text{net}\;} = m(g+y v^{2})$ Acceleration is given by, $a = \frac{F}{m}$ $a = \frac{m(g+\gamma v^{2})}{m}$ $a = -(g+y v^{2})$ Thus, net acceleration is the change in the velocity, $\frac{dv}{dt} = a$ $\frac{dv}{dt} = -(g + \gamma v^{2})$ $\frac{dv}{-(g + \gamma v^{2})} = dt$ $\frac{-dv}{g + \gamma v^{2}} = dt$ Integrating both the sides, $\int\limits_{v_0}^{v} \frac{-dv}{g + \gamma v^{2}} = \int\limits_{0}^{t} dt$ Let time $t$ required to rise to its zenith $(v=0)$ so, $\Rightarrow \frac{1}{\gamma} \int\limits_{v_0}^{0} \frac{-dv}{\frac{g}{\gamma}+v^{2}} = \int\limits_{0}^{t} dt$ We know that, according to integral formula, $\int \frac{1}{x^{2}+a^{2}} dx = \frac{1}{a} \tan^{-1}\left( \frac{x}{a} \right) + C$ Thus, $\Rightarrow \frac{1}{\gamma} \int\limits_{v_0}^{0} \frac{-dv}{\left( \sqrt{\frac{8}{\gamma}} \right)^{2} + (v)^{2}} = \int\limits_{0}^{t} dt$ Here, $x=\sqrt{\frac{g}{\gamma}}, a=v$ $\Rightarrow t = \frac{1}{\gamma} \left\{ \left[ -\frac{1}{\sqrt{\frac{g}{\gamma}}} \tan^{-1}\left( \frac{v}{\sqrt{\frac{g}{\gamma}}} \right) \right]_{V_0}^{0} + C \right\}$ $[$ for $H_{\max}, v=0]$ $t=0 - \frac{1}{\gamma} \left( -\sqrt{\frac{\gamma}{g}} \tan^{-1}\left( \frac{\sqrt{\gamma} V_0}{\sqrt{g}} \right) \right)$ $\Rightarrow t = \frac{1}{\sqrt{\gamma g}} \tan^{-1}\left( \frac{\sqrt{\gamma} V_0}{\sqrt{g}} \right)$ Therefore, the time taken by the ball to rise to its zenith is $\frac{1}{\sqrt{\gamma g}} \tan^{-1}\left( \sqrt{\frac{\gamma}{g}} V_0 \right)$