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Question

At a height of 0.4 m from the ground, the velocity of a projectile is v=(6^i+2^j) m/s . The angle of projection is (g=10 m/s2 ).
  1. 45
  2. 60
  3. 30
  4. tan1(34)


Solution

The correct option is C 30
Given v=vx^i+vy^j=6^i+2^j
As the horizontal component of velocity always remains constant
vx=ux=6
By using third equation of motion;
v2y=u2y2gh
22=u2y2(10)(0.4)
u2y=12 or uy=12=23
tanθ=uyux=13θ=30°
Hence, the correct answer is option (c).

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