Q: The velocity of a projectile at its greatest height is √(2/5) times its velocity, at half of its greatest height, find the angle of projection.

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Sol: $\large u cos\theta = \sqrt{\frac{2}{5}} \times u \sqrt{\frac{1+cos^2\theta}{2}}$

Squaring on both sides

$\large u^2 cos^2 \theta = \frac{2}{5}u^2 (\frac{1+cos^2 \theta}{2}) $

$\large 4 cos^2 \theta = 1 $

$\large cos^2 \theta = \frac{1}{4}$

θ = 60°