question_answer

If a particle is projected with speed u from ground at an angle with horizontal, then radius of curvature of a point where velocity vector is perpendicular to initial velocity vector is given by

A) \[\frac{{{u}^{2}}{{\cos }^{2}}\theta }{g}\]    

B)        \[\frac{{{u}^{2}}{{\cot }^{2}}\theta }{g\sin \theta }\]

C) \[\frac{{{u}^{2}}}{g}\]                       

D)        \[\frac{{{u}^{2}}{{\tan }^{2}}\theta }{g\cos \theta }\]

Correct Answer: B

Solution :

[b] Horizontal components of velocity at O and P are equal. \[\therefore \text{v}\cos \left( 90{}^\circ -\theta  \right)=\text{u}\cos \theta \] \[\text{or }\,\text{v sin}\theta =u\cos \theta \] \[\text{or }\,\text{v}\,\text{=}\,\,\text{u}\,\text{cos}\theta \] \[\text{At }\,\text{P, }\frac{\text{v}_{T}^{2}}{R}={{a}_{c}}\text{ ;}\] \[\frac{{{\text{u}}^{2}}{{\cot }^{2}}\theta }{g\sin \theta }=R\]