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A projectile is fired at an angle of 30° to the horizontal such that the vertical component of its initial velocity is $80 m s^{- 1} .$ Its time of flight is T. Its velocity at $t = \frac{T}{4}$ has a magnitude of nearly

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$200 m s^{- 1}$

$300 m s^{- 1}$

$140 m s^{- 1}$

$100 m s^{- 1}$

answer is C.

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Detailed Solution

$\frac{u_{x}}{u_{y}} = \cot 30^{0} = \sqrt{3}$

$∴ u_{x} = 80 \sqrt{3} m s^{- 1}$

$T = \frac{2 u_{y}}{g} = \frac{2 \times 80}{10} = 16 s$

$A t t = \frac{T}{4} = 4 s, v_{x} = 80 \sqrt{3} m s^{- 1}$

$v_{y} = 80 - 10 \times 4 = 40 m s^{- 1}$

$∴ V = \sqrt{{(80 \sqrt{3})}^{2} + {(40)}^{2}} \approx 140 m s^{- 1}$

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