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Find the angle of projection at which the horizontal range and maximum height of the projectile are equal. |
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Answer» Suppose that θ is the angle of projection at which the horizontal range and maximum height of the projectile become equal. We know, \(H=\frac{u^2sin^2\theta}{2g}\) and \(R=\frac{u^2sin\,2\theta}{g}\) Given H = R \(\frac{u^2sin\,2\theta}{g}=\frac{u^2sin^2\theta}{2g}\) ∵ or sin 2θ = \(\frac{1}{2}sin^2\,\theta\) or 2 sin θ cos θ = \(\frac{1}{2}sin^2\,\theta\) or tan θ = 4 or θ = 75.96º |
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