Question

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. $\frac{v}{2}\sqrt{1+{\cos }^2\theta }$
• C. $\frac{v}{2}\sqrt{1+3{\cos }^2\theta }$
• D. $v\cos \theta$

Answer 1

The correct option is C $\frac{v}{2}\sqrt{1+3{\cos }^2\theta }$

For highest point-

$T=(total-time)/2=\frac{v\sin \theta }{g}$

$x=v\cos \theta \times T=\frac{v^2\sin \theta \cos \theta }{g}$

$y=\frac{v^2{\sin }^2\theta }{2g}$

displacement $s=\sqrt{x^2+y^2}=\frac{v^2}{g}\sqrt{{\sin }^2\theta {\cos }^2\theta +\frac{{\sin }^4\theta }{4}}$

$v_{av}=\frac{s}{T}=v\sqrt{{\cos }^2\theta +\frac{{\sin }^2\theta }{4}}$

$⟹v_{av}=\frac{v}{2}\sqrt{3{\cos }^2\theta +1}$

Answer-(C)