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A piece of marble is projected from earth's surface with velocity of $19.6 \sqrt{2}$ m/s at 45^o. 2 s later its velocity makes an angle $α$ with horizontal, where $α$ is

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45^o

60^o

30^o

0^o

answer is D.

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Detailed Solution

v_x= 19.6 m/s and v_y= (19.6-9.8t) m/s . After 2 s, vertical component of velocity v_ywill become zero.

So, particle is at its highest point. $α$ =0

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