Question

A particle is projected from the ground with an initial speed of v at an angle q 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 cosq

Answer 1

Answer: (C)

$\frac { v } { 2 }$ $\sqrt{1 + 3\cos^{2}\theta}$

Answer 2

Average velocity =$\frac{displacement}{time}$

v_av =$\frac{\sqrt{H^{2} + \frac{R^{2}}{4}}}{T/2}$ … (1)

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

Substituting in Eq. (1) we get

v_av = $\frac { v } { 2 }$ $\sqrt{1 + 3\cos^{2}\theta}$