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

Answers (1)

A avinash.dongre

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 $\theta =tan^{-1}(3)$

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

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