A body has maximum range $R_{1}$ when projected up the inclined plane. The same body when projected down the inclined plane with same velocity it has maximum range $R_{2}$ . The maximum horizontal range for same velocity is

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$\frac{R_{1} R_{2}}{R_{1} + R_{2}}$

$\frac{{R_{2}}^{2}}{R_{1} + R_{2}}$

$\frac{{R_{1}}^{2}}{R_{1} + R_{2}}$

$\frac{2 R_{1} R_{2}}{R_{1} + R_{2}}$

answer is D.

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Detailed Solution

$m a x i m u m r a n g e o n i n c l i n e d p l a n e w h e n p r o j e c t e d u p t h e p l a n e = R_{1} = \frac{u^{2}}{g (1 + \sin β)}; m a x i m u m r a n g e o n i n c l i n e d p l a n e w h e n p r o j e c t e d d o w n t h e p l a n e = R_{2} = \frac{u^{2}}{g (1 - \sin β)} m a x i m u m r a n g e w h e n p r o j e c t e d o n t h e g r o u n d = R = \frac{u^{2}}{g} \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{1}{\frac{u^{2}}{g (1 + \sin β)}} + \frac{1}{\frac{u^{2}}{g (1 - \sin β)}} \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{2 g}{u^{2}} \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{2}{R} R = \frac{2 R_{1} R_{2}}{R_{1} + R_{2}}$