Question

A particle is projected from ground with speed $u$ such that the horizontal range is $R$ and maximum height is $H$. The angular velocity of the particle about the point of projection, when it is at the highest point of its trajectory is:

• A. $\frac{HRu}{(R^2+4H^2)\sqrt{R^2+16H^2}}$
• B. $\frac{4HRu}{(R^2+4H^2)\sqrt{R^2+16H^2}}$
• C. $\frac{u}{H}$
• D. $\frac{uR}{\left(H\right)\sqrt{R^2+16H^2}}$

Answer 1

The correct option is B $\frac{4HRu}{(R^2+4H^2)\sqrt{R^2+16H^2}}$

$H=\frac{u^2{\sin }^2\alpha }{2g}$ ... (1)
$R=\frac{u^2\sin 2\alpha }{g}$ ... (2)
From (1) and (2)
$\tan \alpha =\frac{4H}{R}$
Now, $\omega =\frac{v_{\perp }}{r}=\frac{u\cos \alpha \times \sin \theta }{r}$
$=\frac{u\times R}{\sqrt{16H^2+R^2}}\times \frac{1}{\sqrt{\frac{R^2}{4}+H^2}}\times \frac{H}{\sqrt{\frac{R^2}{4}+H^2}}$
$\omega =\frac{4uRH}{(R^2+4H^2)\sqrt{16H^2+R^2}}$

Why this question? Tip: The result $\tan \alpha =\frac{4H}{R}$ should be remembered and directly used.