Question

A body is projected from the earth at angle $30^{\circ }$ with the horizontal with some initial velocity. If its range is $20m$, the maximum height reached by it is : (in metres)

• A. $5\sqrt{3}$
• B. $\frac{5}{\sqrt{3}}$
• C. $\frac{10}{\sqrt{3}}$
• D. $10\sqrt{3}$

Answer 1

Answer: (B)

$R=\frac{u^2\sin 2\theta }{g}$
$\therefore 20=\frac{u^2\sin \left(2\times 30^{\circ }\right)}{g}$
$\Rightarrow \frac{u^2}{g}=\frac{20}{\sin 60^{\circ }}=\frac{20}{\sqrt{3}}\times 2=\frac{40}{\sqrt{3}}$
Now, $H=\frac{u^2{\sin }^2\theta }{2g}$
$=\frac{40}{\sqrt{3}}\times \frac{{\sin }^{2}30^{\circ }}{2}$
$=\frac{40}{\sqrt{3}}\times \frac{{\left(\frac{1}{2}\right)}^2}{2}$
$=\frac{5}{\sqrt{3}}m$