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A particle is projected with a speed of $10 \sqrt{3} {ms}^{- 1}$ at an angle of 60° from the horizontal. The velocity of the projectile when it reaches a height of 10m is $(g = 10 m / s^{2})$

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5 $\sqrt{3}$ m/s

5 m/s

15 m/s

10 m/s

answer is A.

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Detailed Solution

$u_{x} = 10 \sqrt{3} \cos 60^{0} = 5 \sqrt{3} m / s$

$\Rightarrow v_{x} = u_{x} = 5 \sqrt{3} m / s$

$u_{y} = 10 \sqrt{3} \sin 60^{0} = 15 m / s$

$v_{y} = \sqrt{{u_{x}}^{2} - 2 gh}$

$\Rightarrow v_{y} = \sqrt{15^{2} - 2 \times 10 \times 10} = 5 m / s$

$∴ v = \sqrt{{(5 \sqrt{3})}^{2} + 5^{2}} = 10 m / s$

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