• # question_answer A particle is projected from the ground with an initial speed of v at an angle$\theta$ with horizontal. The magnitude of 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) $v\cos \theta$

Initial velocity $=v\cos \theta \hat{i}+v\sin \theta \hat{j}$ Final velocity $=v\cos \theta \,\hat{i}$ The$x$component of the velocity remains $v\cos \theta ,$so its average value is also $v\cos \theta ,$whereas the y component of the velocity changes from $v\sin \theta$to, zero, so its average value is$\frac{v\sin \theta }{2}.$ $\therefore$Average velocity$=v\cos \theta \,\hat{i}+\frac{v\sin \theta \hat{j}}{2}$ Magnitude of average velocity $=\frac{1}{2}\sqrt{4{{v}^{2}}{{\cos }^{2}}\theta +{{v}^{2}}{{\sin }^{2}}\theta }$ $=\frac{1}{2}\sqrt{{{v}^{2}}+3{{v}^{2}}{{\cos }^{2}}\theta }=\frac{v}{2}\sqrt{1+3{{\cos }^{2}}\theta }$ Hence, the correction option is [c].