# At a height 0.4 m from the ground the velocity of particle projected in angle $\theta$ is $\hat v=(6 i+ 2 j)m/s$ The angle of projection $\theta$ is $(g=10 m/s^2)$

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

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

edited Jul 25, 2014