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A particle of mass m is projected with a velocity u making an angle of $30^{\circ}$ with the horizontal. The magnitude of $(V_{h} \times h)$ of the projectile when the particle is at its maximum height h
Explanation
$V_{h} = V \cos \theta$ where h is the maximum height $V_{h} \times h = (v \cos \theta)\left( \frac{v^{2}\sin^{2} \theta}{2g} \right) = \frac{v^{3}\sin^{2} \theta \cos \theta}{2g} = \sqrt{3}v^{3}/16g$