Q. A body is projected from the earth at angle $30{^\circ}$ with the horizontal with some initial velocity. If its range is 20 m, the maximum height reached by its is (in metre )
A) $5\sqrt{3}$
B) $\frac{5}{\sqrt{3}}$
C) $\frac{10}{\sqrt{3}}$
D) $10\sqrt{3}$
Explanation
(b) $Range,R = \frac{u^{2}\sin{2\theta}}{g}$
$\therefore 20 = \frac{u^{2}\sin(2 \times 30{^\circ})}{g}$
$\Longrightarrow \frac{u^{2}}{g} = \frac{20}{\sin{60{^\circ}}} = \frac{20}{\sqrt{3}} \times 2 = \frac{40}{\sqrt{3}}$
$Now,\ H = \frac{u^{2}\sin^{2}\theta}{2g}$
$= \frac{40}{\sqrt{3}} \times \frac{\sin^{2}30{^\circ}}{2}$
$= \frac{40}{\sqrt{3}} \times \frac{\left( \frac{1}{2} \right)^{2}}{2} = \frac{5}{\sqrt{3}}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.