question_answer

A particle is projected horizontally with muzzle velocity \[v\] from the top of a plane inclined at an angle \[\alpha \] with the horizontal. It strikes the plane at point P shown in the figure. The time taken by particle to reach at point P in terms of u, g,\[\text{ }\!\!\theta\!\!\text{ }\]would be

A)  \[\frac{2v\,\sin \theta }{g\,(\cos \theta )}\]                       

B)  \[\frac{v\,\cos \theta }{(\sin \theta )g}\]

C)  \[\frac{2v}{(\cot \theta -1)}\]                   

D)  \[\frac{1}{3}vg\,\tan \theta \]

Correct Answer: A

Solution :

                Consider the diagram For the motion of particle along X-axis \[AP=(v\,\cos \theta )t+\frac{1}{2}(g\sin \theta ){{t}^{2}}\] For the motion of particle along Y-axis displacement, \[0=-\,vt\,\sin \theta +\frac{1}{2}g{{t}^{2}}\cos \theta \] This gives, \[t=0,\,\frac{2v\,\sin \theta }{g\,\cos \theta }\] Clearly, time corresponding to point P is\[2v\sin \theta /g\cos \theta \]