A stone with weight w is thrown vertically upward into air from ground level with initial speed v₀. If a constant force f due to air drag acts on the stone throughout its flight, the speed of stone just before impact with the ground is

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$v_{0} {(\frac{w + f}{w - f})}^{1 / 2}$

$v_{0} {(\frac{w - f}{w + f})}^{1 / 2}$

$v_{0} {(\frac{w + f}{f})}^{1 / 2}$

$v_{0} {(\frac{w - f}{w})}^{1 / 2}$

answer is A.

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Detailed Solution

The upward motion will have a deceleration of g+f/m

and the downward motion will have an acceleration of g-f/m

$\begin{array}{l} H = \frac{v_{0}^{2}}{2 (g + \frac{f}{m})} = \frac{{mv}_{0}^{2}}{2 (w + f)} \\ V^{2} = 2 (g - \frac{f}{m}) \frac{{mv}_{0}^{2}}{2 (w + f)} = \frac{2 (w - f)}{m} \times \frac{{mv}_{0}^{2}}{2 (w + f)} V = V_{0} \sqrt{\frac{w - f}{w + f}} \end{array}$