Solution:
Comparing with \(y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}\), we find \(\tan\theta = \sqrt{3}\), so horizontal angle \(\theta = 60^\circ\). The angle with vertical is \(90^\circ - 60^\circ = 30^\circ\).
Equation of trajectory of a projectile is \(y = \sqrt{3}x - 5x^2\). Then angle of projection with vertical is (Assume x-axis as horizontal and y-axis as vertical)
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