A particle is projected horizontally as shown in fig. $(g = 10 m / s^{2})$ . The values of time of flight and range are

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The time of flight of the projectile along the inclined plane is 2.31s

Range of the projectile along the inclined plane is 53.33m

The time of flight of the projectile along the inclined plane is 23.1s

Range of the projectile along the inclined plane is 23.33m

The time of flight of the projectile along the inclined plane is 2.99s

Range of the projectile along the inclined plane is 23.33m

The time of flight of the projectile along the inclined plane is 23.1s

Range of the projectile along the inclined plane is 53.33m

answer is A.

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Detailed Solution

$\begin{array}{l} t i m e o f f l i g h t f o r a b o d y t h a t i s p r o j e c t e d d o w n t h e i n c l i n e d p l a n e = T = \frac{2 u \sin (α + β)}{g \cos β} \\ h e r e a n g l e o f p r o j e c t i o n w r t h o r i z o n t a l = α = 0^{\circ}; a n g l e l m a d e b y t h e i n c l i n e d p l a n e w r t h o r i z o n t a l = β = 30^{0}; \\ a c c e l e r a t i o n d u e t o g r a v i t y = g = 10 m / s^{2}; s u b s t i t u t e g i v e n v a l u e s i n a b o v e e q u a t i o n \\ T = \frac{(2) (20) \sin (30^{\circ})}{(10) \cos 30^{\circ}} = 2.31 s \\ r a n g e = R = \frac{u^{2}}{g \cos^{2} β} [\sin (2 α + β) + \sin β]; h e r e u = i n i t i a l v e l o c i t y o f t h e b o d y p r o j e c t e d \\ α = 0^{\circ} \\ R = \frac{[(20)^{2}]] [\sin 30^{\circ} + \sin 30^{\circ}]}{(10) \cos^{2} 30^{\circ}} = 53.33 m \end{array}$