• # question_answer A particle is projected from a point A with a velocity V at an angle $\theta$ (upward) with the horizontal. At a certain point B, it moves at right angle to its initial direction. It follows that: A)  velocity of the particle at B is VB)  velocity of the particle at B is V $\cos \theta$C)  velocity of the particle at B is $V\tan \theta$D)  the time of flight from A to B is $\frac{V}{g\sin \theta }$

Solution :

As,           $\upsilon =u+at$ Considering the motion along the line AC             $0=\upsilon -g\,\sin \theta \,t$                 $t=\frac{\upsilon }{g\sin \theta }$ Now consider the motion along the line C.             $v'=0+g\cos \theta =\frac{V}{g\sin \theta }=\upsilon \cot \theta$

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