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  • If the horizontal range of a projectile is 4/3 times its maximum height. Find the angle of the projectile

If the horizontal range of a projectile is 4/3 times its maximum height. Find the angle of the projectile

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Maximum Height - 

 

H= \frac{U^{2}\sin ^{2}\Theta }{2g}

Horizontal Range

    R=\frac{u^{2}\sin 2\Theta }{g}

So we get

4H = Rtanθ

4H = (4H/3) tanθ
3   = tanθ

So \theta =tan^{-1}(3)

Posted by

avinash.dongre

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