question_answer

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

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

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

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

D) \[v\,\cos \,\theta \]

Correct Answer: C

Solution :

[c] Average velocity \[=\frac{Displacement}{Time}\] \[=\frac{\sqrt{{{H}^{2}}+{{R}^{2}}/4}}{T/2}\] Putting the required values, we get \[{{v}_{av}}=\frac{v}{2}\sqrt{1+3{{\cos }^{2}}\theta }\]