At a height 0.4 m from the ground, the velocity of a projectile in vector form is v→=(6i^+2j^)ms−1 The angle of projection is

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45o

60o

30o

tan-1 (3/4)

answer is C.

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Detailed Solution

v2=u2-2gh or u2=v2+2gh ux2+uy2=vx2+vy2+2gh,ux=vx uy2=vy2+2gh or uy2=(2)2+2×10×0.4=12 uy=12=23 ms-1,ux=vx=6 ms-1 tanθ=uyux=236=13 ⇒ θ=30°

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