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 :

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