A ball is thrown upward with an initial veloc | vedclass

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Question

(A) During the upward motion,the forces acting on the ball are gravity $(mg)$ acting downwards and the drag force $(m \gamma v^2)$ also acting downwar

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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