Question

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

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

Answer 1

Answer: (A)

$\sqrt{3}u^{2}/2g$

Answer 2

If the velocity of projection is u then at the highest point body posses only $u\cos\theta$

$u\cos\theta = \frac{u}{2}$ (given) $\therefore\theta = 60{^\circ}$

Now $R = \frac{u^{2}\sin(2 \times 60{^\circ})}{g}$ $= \frac{\sqrt{3}}{2}\frac{u^{2}}{g}$