A small marble is projected with a velocity of $10 m / s$ in a direction $45 °$ from the $y -$ direction of the smooth inclined plane, Calculate the magnitude $v$ of its velocity after $2 s$ . (Take, $g = 10 m / s^{2}$ )

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10 m/ s

30 m/ s

70 m/ s

$g \sin 45 °$

answer is D.

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Detailed Solution

It is just like a projectile motion with g to be replaced by $g \sin 45 °$ .

After 2 s,

$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{{(10 \sin 45 ° - \frac{g}{\sqrt{2}} \times 2)}^{2} + {(10 \cos 45 °)}^{2}} = 10 m / s$

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