question_answer

A particle is projected from the ground with an initial speed of\[v\]at an angle\[\theta \]with horizontally. 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 :

 \[\text{Average}\,\text{velocity}\,\,\text{=}\,\,\frac{\text{Displacement}}{\text{Time}}\] \[v=\frac{\sqrt{{{H}^{2}}+\frac{{{R}^{2}}}{4}}}{\frac{T}{2}}\] Here, H = maximum height\[=\frac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}\] \[R=range=\frac{{{v}^{2}}\sin 2\theta }{g}\] and \[T=\text{time of light}=\frac{2v\sin \theta }{g}\] \[\therefore \] \[{{v}_{av}}=\frac{v}{2}\sqrt{1+3{{\cos }^{2}}\theta }\]