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The velocity of a projectile when it is ...

The velocity of a projectile when it is at the greatest height is `(sqrt (2//5))` times its velocity when it is at half of its greatest height. Determine its angle of projection.

Text Solution

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`:. U cos theta=sqrt((2)/(5))xxusqrt((1+cos^(2)theta)/(2))`
Squaring on both sides
`u^(2) cos^(2) theta=(2)/(5) u^(2)((1+cos^(2) theta)/(2))`
`10 cos^(2) theta=2+2cos^(2) theta`
`rArr 8 cos^(2) theta=2 rArr cos^(2) theta=(1)/(2) rArr theta=60^(@)`
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