The ratio of radius of curvature at the highest point and the point of projection of a projectile is 1/8. Find the angle of projection in degrees. ( $g = 10 m / s^{2}$ )

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Detailed Solution

$\begin{array}{l} R_{h i g h e s t p o int} = \frac{{(μ \cos θ)}^{2}}{g} \\ R_{i n i t i a l} = \frac{μ^{2}}{g \cos θ} \\ r a t i o = \cos^{3} θ \end{array}$

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