\[(a)\;45 ^{\circ}\quad (b)\;30 ^{\circ} \quad (c)\;60^{\circ} \quad (d)\tan ^{-1} \frac{3}{4}\]

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Let $u_x \;and\; u_y$ be initial component of velocity at the time of projection.

$v=6i+2j$

$\tan \theta=\Large\frac{u_y}{u_x}$

$=>v_x=6;v_y=2$

$=>u_x=v_x$

$=>u_x=6$

${v_y}^2={u_y}^2-2gh$

${u_y}^2={v_y}^2+2gh$

${u_y}^2=(2)^2+2(10)(0.4)=12$

$u_y=\sqrt {12}=2 \sqrt 3$

Therefore $\tan {\theta}=\large\frac{2 \sqrt 3}{6}$

$=>\large\frac{1}{\sqrt 3}$

$\theta=30 ^{\circ}$

Hence b is the correct answer

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