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A particle is projected from the ground with an initial speed of ' $v$ ' at angle $\theta$ with horizontal. The average velocity of the particle between its point of projection and height point of trajectory is
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The correct answer is:
$\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}$
From figure,average velocity,
$v_{\mathrm{av}}=\frac{\sqrt{H^2+R^2 / 4}}{T / 2}$$\ldots(\mathrm{i})$
Here, $H=\frac{u^2 \sin ^2 \theta}{2 g}$
$R=\frac{u^2 \sin 2 \theta}{g}$ and $T=\frac{2 u \sin \theta}{g}$
Putting these value in (i), we get
$v_{\mathrm{av}}=\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}$
$v_{\mathrm{av}}=\frac{\sqrt{H^2+R^2 / 4}}{T / 2}$$\ldots(\mathrm{i})$
Here, $H=\frac{u^2 \sin ^2 \theta}{2 g}$
$R=\frac{u^2 \sin 2 \theta}{g}$ and $T=\frac{2 u \sin \theta}{g}$
Putting these value in (i), we get
$v_{\mathrm{av}}=\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}$
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