At a height of 0.4m from the ground, the velocity of a projectile is →v=(6^i+2^j)m/s . The angle of projection is (g=10m/s2 ).
A
45∘
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B
60∘
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C
30∘
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D
tan−1(34)
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Solution
The correct option is C30∘ 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=u2y−2gh 22=u2y−2(10)(0.4) ⇒u2y=12 or uy=√12=2√3 tanθ=uyux=1√3⇒θ=30°
Hence, the correct answer is option (c).