A projectile is projected at 30° from horizontal with initial velocity $40 {ms}^{- 1}$ . The velocity of the projectile at $t = 2 s$ from the start will be: (Given $g = 10 m / s^{2}$ )

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$20 \sqrt{3} {ms}^{- 1}$

$20 {ms}^{- 1}$

$40 \sqrt{3} {ms}^{- 1}$

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answer is A.

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Detailed Solution

$u_{x} = 40 \cos 30^{o} = 20 \sqrt{3}$

$u_{y} = 40 \sin 30^{o} = 20$

$v_{x} = u_{x}, v_{y} = u_{y} - gt$

at t=2 sec
$v_{x} = u_{x} = 20 \sqrt{3}$

$v_{y} = 20 - 10 (2) = 0$

$\Rightarrow v = \sqrt{v_{x}^{2} + v_{y}^{2}} = 20 \sqrt{3} {ms}^{- 1}$

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