Question

A particle is projected at an angle of $60^\circ $ above the horizontal with a speed of $10 \, m \, s^{- 1}$ . After some time the direction of its velocity makes an angle of $30^\circ $ above the horizontal. The speed of the particle at this instant is

• A. $\frac{5}{\sqrt{3}}ms^{- 1}$
• B. $5\sqrt{3} \, m \, s^{- 1}$
• C. $5 \, m \, s^{- 1}$
• D. $\frac{10}{\sqrt{3}}ms^{- 1}$

Answer 1

Answer: (D)

Let $v$ be the velocity of point where it make angle $\text{30}^{\text{ο}}$ with
horizontal
thus , horizontal velocity remains uncharged, so

$v cos 30^{\text{ο}} = 10 cos 60^{\text{ο}}$
$v \frac{\sqrt{3}}{2} = \frac{\text{10}}{2}$
$v=\frac{\text{10}}{\sqrt{3}}ms^{- 1}$