Question

A particle is projected from the ground with an initial speed of u at an angle $\theta$ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is :-

• A. $\frac{u}{2}\sqrt{1+2cos\theta }$
• B. $\frac{u}{2}\sqrt{4-3si{n}^2\theta }$
• C. $\frac{u}{2}\sqrt{1+3co{s}^2\theta }$
• D. $ucos\theta$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{u}{2}\sqrt{4-3si{n}^2\theta }$
C $\frac{u}{2}\sqrt{1+3co{s}^2\theta }$

$<v_{avgx}>=\frac{u\cos \theta +u\cos \theta }{2}=u\cos \theta$

$<v_{avgy}>=\frac{u\sin \theta +0}{2}=\frac{u\sin \theta }{2}$

$v_{avg}=\sqrt{<v_{avgx}{>}^2+<v_{avgy}{>}^2}$

$v_{avg}=\sqrt{u^2{\cos }^2\theta +(\frac{u\sin \theta }{2}{)}^2}$

$v_{avg}=u\sqrt{{\cos }^2\theta +\frac{{\sin }^2\theta }{4}}$

$v_{avg}=\frac{u}{2}\sqrt{1+3{\cos }^2\theta }$

$v_{avg}=\frac{u}{2}\sqrt{1+3(1-{\sin }^2\theta )}$

$v_{avg}=\frac{u}{2}\sqrt{4-3{\sin }^2\theta )}$