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A particle is projected from the ground ...

A particle is projected from the ground with an initial speed of u at an angle of projection `theta`. The average velocity of the particle reaches highest point of trajectory is

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`v_("avg") =(vecv + vecu)/2 = (u cos thetahati +(u cos thetahati + u sintheta))/2`
`v_(av) =v/2sqrt(1+3 cos^(2)theta)`
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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