Question

Average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory is (projection speed$=u$, angle of projection form horizontal $=\theta$)

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

Answer 1

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

Displacement = $\sqrt{(\frac{R}{2}{)}^2+H^2}$

$R=\frac{u^2\sin 2\theta }{g}$

$H=\frac{u^2{\sin }^2\theta }{g}$

$t=\frac{T}{2}=\frac{u\sin \theta }{g}$

$|V_{avg}|=\frac{|displacement|}{t}=\frac{u^2\sin \theta }{g}\sqrt{4{\cos }^2\theta +{\sin }^2\theta }\sqrt{\frac{1}{4}}\frac{1}{\frac{u\sin \theta }{g}}=\frac{u}{2}\sqrt{1+3{\cos }^2\theta }$