1.

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

Answer»

Solution :
`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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