Average velocity of a particle in projectile | vedclass

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(B) The displacement vector $\vec{s}$ from the starting point $(0,0)$ to the highest point $(\frac{R}{2}, H)$ is given by $\vec{s} = \frac{R}{2} \hat{

Vedclass · Accepted Answer

$\frac{u}{2} \sqrt{1 + 3 \cos^2 	heta}$

Vedclass · Answer

(B) The displacement vector $\vec{s}$ from the starting point $(0,0)$ to the highest point $(\frac{R}{2}, H)$ is given by $\vec{s} = \frac{R}{2} \hat{i} + H \hat{j}$.The magnitude of displacement is $|\vec{s}| = \sqrt{(\frac{R}{2})^2 + H^2}$.Given $R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$ and $H = \frac{u^2 \sin^2 	heta}{2g}$.So,$\frac{R}{2} = \frac{u^2 \sin 	heta \cos 	heta}{g}$.$|\vec{s}| = \sqrt{(\frac{u^2 \sin 	heta \cos 	heta}{g})^2 + (\frac{u^2 \sin^2 	heta}{2g})^2} = \frac{u^2 \sin 	heta}{g} \sqrt{\cos^2 	heta + \frac{\sin^2 	heta}{4}} = \frac{u^2 \sin 	heta}{2g} \sqrt{4 \cos^2 	heta + \sin^2 	heta} = \frac{u^2 \sin 	heta}{2g} \sqrt{3 \cos^2 	heta + 1}$.The time taken to reach the highest point is $t = \frac{u \sin 	heta}{g}$.Average velocity $V_{avg} = \frac{|\vec{s}|}{t} = \frac{\frac{u^2 \sin 	heta}{2g} \sqrt{3 \cos^2 	heta + 1}}{\frac{u \sin 	heta}{g}} = \frac{u}{2} \sqrt{1 + 3 \cos^2 	heta}$.

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 from horizontal = $\theta$)

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