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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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Step 1: we know that, velocity of a projectile at half of maximum height: `=usqrt((1+cos^(2)theta)/2)`
Step 2: given that `u cos theta =sqrt(2/5) xx u sqrt((1+cos^(2)theta)/2)`
Squaring on both sides `u^(2) cos^(2)theta =2/5u^(2)(1+cos^(2)theta)/2`
`10cos^(2)theta =2 + 2 cos^(2)theta rArr theta=60^(@)`
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