question_answer

A particle is projected from a point O with a velocity u in a direction making an angle \[\theta \]  upward with the horizontal. After some time at point P it is moving at right angle with its initial direction of projection. The time of flight from O to P is

A) \[\frac{u\,\sin \,\alpha }{g}\]                   

B) \[\frac{u\,\,\cos ec\,\alpha }{g}\]

C) \[\frac{u\,\tan \,\alpha }{g}\]                  

D) \[\frac{u\,\sec \,\alpha }{g}\]

Correct Answer: B

Solution :

When body projected with initial velocity \[\overrightarrow{u}\] by making angle \[\alpha \] with the horizontal. Then after time t, (at point P) its direction is perpendicular to\[\overrightarrow{u}\]. Magnitude of velocity at point P is given by \[\operatorname{v} = u\,\,cot\,\alpha .\] For vertical motion: Initial velocity (at point O) \[= u\,\,\sin \,\alpha .\] Final velocity (at point P)                    \[= - v cos\,\alpha  = -u\,cot\,\alpha \,cos \alpha \] \[\operatorname{Time} of flight \left( from pointOto P \right) =t\] Applying first equation of motion \[\operatorname{v} =u-gt\] \[-u\,cot \alpha  cos \alpha  =u\,sin\,\alpha  - gt\] \[\therefore \,\,\,t=\frac{u\,\,\sin \,\alpha \,+u\cot \,\alpha \,\cos \,\,\alpha }{g}~~~~~~~~~~\] \[=\,\,\frac{u}{g\,\,\sin \,\,\alpha }\,\left[ {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha  \right]\,\,=\,\,\frac{u\,\,\operatorname{cosec}\,\alpha }{g}\]