Q123 / 160 · AP EAMCET 2006 · Physics

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

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

Solution

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

Hence, the correct answer is option: 'B'.

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