Question

Find the angle of projection for a projectile motion whose range r is n times the maximum height

Answer 1

Answer: tan-1(4/n)

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Answer 2

The angle of projection of the projectile is $\tan^{-1}(\dfrac{4}{n})$.

Explanation:

The range R and maximum height H reached by the projectile is given by :

$R=\dfrac{u^2\sin 2\theta}{g}\\\\H=\dfrac{u^2\sin ^2\theta}{2g}$

According to given condition,

$R=nH\\\\\dfrac{u^2\sin 2\theta}{g}=n\times \dfrac{u^2\sin ^2\theta}{2g}\\\\\sin 2\theta=n\times \dfrac{\sin ^2\theta}{2}\\\\2\sin \theta cos \theta=n\times \dfrac{\sin ^2\theta}{2}\\\\4\cos\theta=n\sin \theta\\\\\tan\theta=\dfrac{4}{n}\\\\\theta=\tan^{-1}(\dfrac{4}{n})$

So, the angle of projection of the projectile is $\tan^{-1}(\dfrac{4}{n})$.

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Projectile motion

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