An aeroplane takes off at an angle of $30^{o}$ to the horizontal. If the component of its velocity along the horizontal is $250 k m - h^{- 1}$ , what is its actual velocity?

314.6 k m - h^{- 1}

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288.7 k m - h^{- 1}

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208.5 k m - h^{- 1}

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128.6 k m - h^{- 1}

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Solution

The correct option is B $288.7 k m - h^{- 1}$
Given that,
Angle made by plane's flight with horizontal $= 30^{\circ}$

Let plane's velocity be $V$

$\Rightarrow$ Horizontal Velocity $= V c o s 30^{\circ} = 250$ km/hr

$\Rightarrow V \frac{\sqrt{3}}{2} = 250$

$∴ V = \frac{500}{\sqrt{3}} = 288.7$ km/hr

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