Question

At a height $0.4m$ from the ground, the velocity of a projectile in vector form is $\overrightarrow{v}=(6\hat{i}+2\hat{j})m{s}^{-1}$. The angle of projection is

• A. ${45}^{\circ }$
• B. ${60}^{\circ }$
• C. ${30}^{\circ }$
• D. $\tan -1\left(3/4\right)$

Answer 1

Answer: (C)

${30}^{\circ }$

$v^2=u^2-2gh$ or $u^2=v^2+2gh$
or $u_x^2+u_y^2+v_x^2+v_y^2+2gh,u_x=v_x$
So, $u_y^2=v_y^2+2gh$ or $u_y^2=(2{)}^2+2\times 10\times 0.4=12$
$u_y=\sqrt{12}=2\sqrt{3}m{s}^{-1},u_x=v_x=6m{s}^{-1}$
$\tan \theta =\frac{u_y}{u_x}=\frac{2\sqrt{3}}{6}=\frac{1}{\sqrt{3}}\Rightarrow \theta ={30}^{\circ }$.