Question

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

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

Answer 1

Answer: (C)

We know, average velocity
$=\frac{\text{ displacement }}{\text{ time }}$
$v_{av}=\frac{\sqrt{{\mu }^2+R^2/4}}{T/2}$ ...(i)

where, $H=$ maximum height
$=\frac{v^2{\sin }^2\theta }{2g}$ ... (ii)
Range $R=\frac{v^2\sin 2\theta }{g}$ ...(iii)
Time of flight $T=\frac{2v\sin \theta }{g}$
Putting the values of Eqs. (ii), (iii) and (iv) in Eq. (i) we have
$v_{av}=\frac{v}{2}\sqrt{1+3{\cos }^2\theta }$