A particle is projected at an angle of 60 ° above the horizontal...

A particle is projected at an angle of

60 °

above the horizontal with a speed of

10ms^{- 1}

. After some time the direction of its velocity makes an angle of

30 °

above the horizontal. The speed of the particle at this instant is A particle is projected at an angle of 60 ° above the horizontal with a speed of 10ms^ - 1 . After some time the

A particle is projected at an angle of 60 ° above the horizontal with a speed of 10ms^ - 1 . After some time the

Options

A $\dfrac{5}{3}ms^{- 1}$
B $5\sqrt{3}ms^{- 1}$
C $5ms^{- 1}$
D $\dfrac{10}{3}ms^{- 1}$

Correct answer

D. $\dfrac{10}{3}ms^{- 1}$

Step-by-step solution

Let

v

be the velocity of point where it make angle

{30}^{ο}

with
horizontal
thus , horizontal velocity remains uncharged, so
A particle is projected at an angle of 60 ° above the horizontal with a speed of 10ms^ - 1 . After some time the

v cos {30}^{ο} = 10 cos {60}^{ο}

v \dfrac{\sqrt{3}}{2} = \dfrac{\text{10}}{2}

v = \dfrac{10}{3}ms^{- 1}

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