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A particle is projected from the ground ...

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

Text Solution

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Average velocity `=("displacement")/("time")`
`V_(av)=(sqrt(H^(2)+(R^(2))/4))/(T/2)`…..i

Here H=maximum height `=(v^(2)sin^(2)theta)/(2g)`
R=range `=(v^(2)sin 2 theta)/g` and `T=` time of flight `=(2vsin theta)/g`
Substituting i (i) we get
`v_(av)=v/2sqrt(1+3 cos^(2)theta)`
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