Question

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 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}{2 g} $
$=\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 $