At a height 0.4 m from the ground, the velocity of a projectile in vector form is : v→=(6i^+2j^)m/s . The angle of projection is :

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450

600

300

tan−1(34)

answer is C.

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Detailed Solution

v2=u2−2gh or u2=v2+2gh or ux2+uy2=vx2+vy2+2gh As vx=ux;uy2=vy2+2gh or uy2=(2)2+2×10×0.4=12 uy=12=23 m/s ux=vx=6 m/s tanθ=uyux=236=13 θ=300 .

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