A) \[{{30}^{{}^\circ }}+{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]
B) \[{{45}^{0}}\]
C) \[{{60}^{0}}\]
D) \[{{30}^{0}}+{{\tan }^{-1}}(2\sqrt{3})\]
Correct Answer: A
Solution :
[a] \[{{t}_{AB}}\]=time of flight of projectile \[=\frac{2u\sin (\alpha -30{}^\circ )}{g\cos 30{}^\circ }\] Now component of velocity along the plane becomes Zero at point B. \[0=u\cos (\alpha -30{}^\circ )-gsin30{}^\circ \times T\] Or \[u\cos (\alpha -30{}^\circ )-gsin30{}^\circ \times \frac{2u\sin (\alpha -30{}^\circ )}{g\cos 30{}^\circ }\] Or \[\tan (\alpha -30{}^\circ )=\frac{\cot 30{}^\circ }{2}=\frac{\sqrt{3}}{2}\] Or \[\alpha =30{}^\circ +{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]You need to login to perform this action.
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