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A particle is projected from the ground with an initial speed of v at an angle of projection $\theta$. The average velocity of the particle between its time of projection and time it reaches highest point of trajectory is

\[(a)\;\frac{v}{2} \sqrt {1+ 2 \cos^2 \theta} \quad (b)\;\frac{v}{2} \sqrt {1+ 2 \sin ^2 \theta} \quad (c)\;\frac{v}{2} \sqrt {1+ 3 \cos^2 \theta} \quad (d)\;v \cos \theta \]

how to (c) answer
plz tell me fast
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1 Answer

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$(c)\;\frac{v}{2} \sqrt {1+ 3 \cos^2 \theta}$
answered Nov 7, 2013 by pady_1

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