Question

A particle is projected from ground with velocity 20 m/s making an angle of 60^o with horizontal. Then angular velocity of the position vector of the particle just before it strikes the ground is

Moderate

Solution

$Range R = \frac{u^{2} \sin 2 α}{g}$
Position vector $\vec{r} = \vec{A B}$
where $r = A B = R = \frac{u^{2} \sin 2 α}{g}$
$∴ ω = \frac{component of velocity at B perpendicular to AB}{AB}$
$= \frac{u \sin α}{R} = \frac{u \sin α}{u^{2} \sin 2 α / g} = \frac{g}{2 u \cos α}$
$∴ ω = \frac{10}{2 \times 20 \times \cos 60^{o}} rad / s = 0.5 rad / \sec$

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