Solution:
Maximum height \(H = \frac{u^2\sin^2\theta}{2g}\) and range \(R = \frac{2u^2\sin\theta\cos\theta}{g}\). Equating the two yields \(\frac{\sin^2\theta}{2} = 2\sin\theta\cos\theta\), which simplifies to \(\tan\theta = 4\).
Maximum height \(H = \frac{u^2\sin^2\theta}{2g}\) and range \(R = \frac{2u^2\sin\theta\cos\theta}{g}\). Equating the two yields \(\frac{\sin^2\theta}{2} = 2\sin\theta\cos\theta\), which simplifies to \(\tan\theta = 4\).
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