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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 ).

A
45
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B
60
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C
30
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D
tan1(34)
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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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