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A projectile is launched with an initial velocity v0 = (2m/s)i + (3m/s)j. At the top of the trajectory, the speed of the particle is (x horizontal direction, y-vertical direction)-

(A) √(22 + 32)m/s

(B) 2m/s 

(C) 3m/s 

(D) 5m/s

1 Answer

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Correct option (B) 2m/s


It maximum height vertical velocity = 0

So horizontal velocity = 2m/s

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