Average velocity of a particle is projectile motion between its starting point and highest point of

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3-2.Motion in Plane

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

$u\,\, cos\theta$

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

$\frac{u}{2}\sqrt {2 + {{\cos }^2}\theta }$

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

Solution

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

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

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

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

$\left|V_{a v g}\right|=\frac{| \text {displacement} |}{t}=\frac{u^{2} \sin \theta}{g} \sqrt{4 \cos ^{2} \theta+\sin ^{2} \theta} \sqrt{\frac{1}{4}} \frac{1}{\frac{u \sin \theta}{g}}$

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

Standard 11

Physics

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(AIIMS-2013)

A particle of mass $m$ is projected with a velocity $V$ making an angle of $45^o$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

normal

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

Solution

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