Question

A particle is projected at an angle of ${60}^o$ above the horizontal with a sped of $10m/s$. After some time, the direction of its velocity makes an angle of ${30}^o$ above the horizontal. The speed of the particle at this instant is

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

Answer 1

The correct option is D $\frac{10}{\sqrt{3}}m/s$

Velocity along $x$ axis remains constant.

Let velocity along $y$ axis at that instant be $v$.

So, according to the question

$\tan {30}^{\circ }=\frac{v}{10\cos {60}^{\circ }}=\frac{v}{5}$

$\frac{1}{\sqrt{3}}=\frac{v}{5}v=\frac{5}{\sqrt{3}}$

$speed=\sqrt{5^2+{\left(\frac{5}{\sqrt{3}}\right)}^2}=\sqrt{25+\frac{25}{3}}$

$=\sqrt{\frac{100}{3}}$

$=\frac{10}{\sqrt{3}}m/s$.