Find the average velocity of a projectile between the instants it crosses one third of the maximum height. It is projected with u making an angle θ with the vertical.
A
ucosθ
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B
u
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C
usinθ
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D
utanθ
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Solution
The correct option is Cusinθ
The vertical velocities at the same height are in opposite directions and therefore their average sum =0. It is horizontal velocity which is uniform and hence vavg=usinθ(=ux).