An object has maximum range ‘X’ when projected up on the inclined plane of angle of inclination $θ$ . The same object projected down the inclined plane has maximum range ‘Y’. If it is projected on horizontal surface with the same angle of projection range is ‘Z’ , initial velocity is same, then the relation between X,Y,Z is

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$Z = X + Y$

$Z = \frac{X + Y}{2}$

$Z = \frac{2 X Y}{X + Y}$

$Z = X Y$

answer is C.

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Detailed Solution

$m a x i m u m r a n g e u p t h e p l a n e i s R = X = \frac{V_{o}^{2}}{g (1 + \sin β)} m a x i m u m r a n g e d o w n t h e p l a n e i s R = Y = \frac{V_{o}^{2}}{g (1 - \sin β)} m a x i m u m r a n g e i n h o r i z o n t a l p l a n e i s R = Z = \frac{V_{o}^{2}}{g} c o n s i d e r t h e i r i n v e r s i o n s, \frac{1}{X} = \frac{g (1 + \sin β)}{V_{o}^{2}}; \frac{1}{Y} = \frac{g (1 - \sin β)}{V_{o}^{2}}; \frac{1}{Z} = \frac{g}{V_{o}^{2}} \frac{2}{Z} = \frac{1}{X} + \frac{1}{Y} ∴ Z = \frac{2 X Y}{X + Y}$