question_answer

A particle is projected from the ground with an initial speed of v at angle \[\theta \]with horizontal. The average velocity of the particle between its point of projection and the highest point of trajectory is

A) \[\frac{v}{2}\sqrt{1+2{{\cos }^{2}}\theta }\]

B)   \[\frac{v}{2}\sqrt{1+{{\cos }^{2}}\theta }\]

C) \[\frac{v}{2}\sqrt{1+3{{\cos }^{2}}\theta }\]

D) \[v\cos \theta \]

Correct Answer: C

Solution :

[c] : From figure, Average velocity, \[{{v}_{av}}=\frac{\sqrt{{{H}^{2}}+({{R}^{2}}/4)}}{T/2}\] (i) Here,\[H=\frac{{{v}^{2}}{{\sin }^{2}}\theta }{2g};\] \[R=\frac{{{v}^{2}}\sin 2\theta }{g}\]and\[T=\frac{2v\sin \theta }{g}\] Putting these values in equation (i), we get, \[{{v}_{av}}=\frac{2}{2}\sqrt{1+3{{\cos }^{2}}\theta }\]