Question

The velocity at the maximum height of a projectile is half its initial velocity of projection $u$ Its range on the horizontal plane is

• A. $ \frac{\sqrt{3}u^{2}}{2g} $
• B. $ \frac{u^{2}}{3g} $
• C. $ \frac{3u^{2}}{2g} $
• D. $ \frac{3u^{2}}{g} $

Answer 1

Answer: (A)

There is only horizontal component of velocity at the highest point because at highest point vertical component of velocity is zero, body begins to come downwards from the highest point.
$\therefore u \cos \theta=\frac{u}{2} \cos \theta$
$=\frac{1}{2}=\cos 60^{\circ}$
$\Rightarrow \theta=60^{\circ}$
ange $R=\frac{2 u^{2} \sin \theta \cos \theta}{g}$
$R=\frac{2 u^{2} \times \sin 60^{\circ} \cos 60^{\circ}}{g}$
$R=\frac{\sqrt{3} u^{2}}{2 g}$