Two stones are projected simultaneously with equal speeds from a point on an inclined plane along the line of its greatest slope upwards and downwards respectively. The maximum distance between their points of striking the plane is double that of when they are projected on a horizontal ground with same speed. If one strikes the plane, two seconds later, the other strikes plane, then

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the angle of inclination of plane is $45 °$

the speeds of their projection is 12.8 m/s

the angle of inclination of plane is 45degree

the speeds of their projection is 11.8 m/s

the angle of inclination of plane is $60 °$

the speeds of their projection is 11.8 m/s

the speeds of their projection is 128 m/s

the angle of inclination of plane is 45 degree

answer is A.

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Detailed Solution

$m a x i m u m r a n g e u p t h e p l a n e = R_{1} = \frac{u^{2}}{g (1 + \sin β)} - - - (1) m a x i m u m r a n g e d o w n t h e p l a n e = R_{2} = \frac{u^{2}}{g (1 - \sin β)} - - - (2) m a x i m u m r a n g e o n g r o u n d = R = \frac{u^{2}}{g} - - - (3) G i v e n R_{1} + R_{2} = 2 (R + R) = 4R s u b s t i t u t e e q u a t i o n s (1), (2), (3) \frac{u^{2}}{g (1 + \sin β)} + \frac{u^{2}}{g (1 - \sin β)} = 4 \frac{u^{2}}{g} o n s o l v i n g w e g e t \cos β = \frac{1}{\sqrt{2}} \Rightarrow β = 45 - - - (4) (b) T i m e o f f l i g h t = T T_{u p t h e p l a n e} = \frac{2 u \sin (α_{1} - β)}{g \cos β} = T_{1} f o r m a x i m u m r a n g e α_{1} = \frac{β}{2} + \frac{π}{4} s u b s t i t u t e i n T_{1} T_{1} = \frac{2 u \sin (\frac{β}{2} + \frac{π}{4} - β)}{g \cos β} = \frac{2 u \sin (\frac{π}{4} - \frac{β}{2})}{g \cos β} - - - (5) T_{d o w n t h e p l a n e} = \frac{2 u \sin (α_{2} + β)}{g \cos β} = T_{2} f o r m a x i m u m r a n g e α_{2} = \frac{π}{4} - \frac{β}{2} T_{2} = \frac{2 u \sin (\frac{π}{4} - \frac{β}{2} + β)}{g \cos β} = \frac{2 u \sin (\frac{π}{4} + \frac{β}{2})}{g \cos β} - - - (6) g i v e n T_{2} - T_{1} = 2 s u b s t i t u t e e q u a t i o n s (5), (6) \frac{2 u \sin (\frac{π}{4} + \frac{β}{2})}{g \cos β} - \frac{2 u \sin (\frac{π}{4} - \frac{β}{2})}{g \cos β} = 2 \Rightarrow u = \frac{g \cos β}{\sin (\frac{π}{4} + \frac{β}{2}) - \sin (\frac{π}{4} - \frac{β}{2})} s u b s t i t u t e e q u a t i o n (4) u = \frac{g \cos \frac{π}{4}}{\sin \frac{3 π}{8} - \sin \frac{π}{8}} = 12.8 m / s$