Q170 / 180 · KCET 2013 · Physics
A projectile is projected at $10 \mathrm{~ms}^{-1}$ by making at an angle $60^{\circ}$ to the horizontal. After some time its velocity makes an angle of $30^{\circ}$ to the horizontal. Its speed at this instant is
- Correct answerA$\frac{10}{\sqrt{3}}$
- B$10 \sqrt{3}$
- C$\frac{5}{\sqrt{3}}$
- D$5 \sqrt{3}$
Solution
As the velocity makes an angle of $30^{\circ}$ with horizontal, so, the horizontal component of velocity at the instant will be $v \cos 30^{\circ}$.
$\begin{array}{ll}
\Rightarrow & v \cos 30^{\circ}=5 \\
\Rightarrow & v=\frac{5}{\cos 30^{\circ}}=\frac{5}{\sqrt{3} / 2}=\frac{10}{\sqrt{3}}
\end{array}$
Hence, the correct answer is option: 'A'.