A ball is thrown upward with an initial veloc | vedclass

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(B) During the upward motion,the forces acting on the ball are gravity $(mg)$ and drag force $(m \gamma v^2)$,both acting downwards.Applying Newton's

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) During the upward motion,the forces acting on the ball are gravity $(mg)$ and drag force $(m \gamma v^2)$,both acting downwards.Applying Newton's second law: $m \frac{dv}{dt} = -(mg + m \gamma v^2)$.This simplifies to: $\frac{dv}{dt} = -(g + \gamma v^2)$.Rearranging the terms for integration: $dt = -\frac{dv}{g + \gamma v^2} = -\frac{1}{\gamma} \frac{dv}{\frac{g}{\gamma} + v^2}$.Integrating from $t = 0$ to $T$ and $v = V_0$ to $0$:$\int_{0}^{T} dt = -\frac{1}{\gamma} \int_{V_0}^{0} \frac{dv}{(\sqrt{g/\gamma})^2 + v^2}$.Using the standard integral $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a})$:$T = \frac{1}{\gamma} \left[ \frac{1}{\sqrt{g/\gamma}} 	an^{-1} \left( \frac{v}{\sqrt{g/\gamma}} ight) ight]_{0}^{V_0}$.$T = \frac{1}{\gamma \sqrt{g/\gamma}} 	an^{-1} \left( V_0 \sqrt{\frac{\gamma}{g}} ight)$.$T = \frac{1}{\sqrt{\gamma g}} 	an^{-1} \left( \sqrt{\frac{\gamma}{g}} V_0 ight)$.

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