A particle is projected from the ground with an initial speed of v at an angle θ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is
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a
v21+2cos2θ
b
v22cos2θ
c
v21+3cos2θ
d
v cos θ
answer is C.
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Detailed Solution
Average velocity =DisplacementTime=H2+R2/4T/2Putting the required values, we get vav=v21+3cos2θ