A projectile is thrown horizontally with a speed of $20 m s^{- 1}$ from certain height. If g is $10 m s^{- 2},$ then the speed of the projectile after 5 second will be nearly

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$0.5 m s^{- 1}$

$5 m s^{- 1}$

$54 m s^{- 1}$

$500 m s^{- 1}$

answer is C.

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Detailed Solution

Even after 5 second, the horizontal velocity $V_{x}$ will be $20 m s^{- 1}$ . The vertical velocity $V_{y}$ is given by $V_{y} = 0 + 10 \times 5 = 50 m s^{- 1}$ Now, $V = \sqrt{V_{x}^{2} + V_{y}^{2}} = \sqrt{20^{2} + 50^{2}} = 54 m s^{- 1}$

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