Question

At a height of $0.4\text{m}$ from the ground, the velocity of a projectile is $\overrightarrow{v}=(6\hat{i}+2\hat{j})\text{m/s}$ . The angle of projection is ($g=10{\text{m/s}}^2$ ).

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

Answer 1

The correct option is C ${30}^{\circ }$

Given $v=v_x\hat{i}+v_y\hat{j}=6\hat{i}+2\hat{j}$
As the horizontal component of velocity always remains constant
$\therefore v_x=u_x=6$
By using third equation of motion;
$v_y^2=u_y^2-2gh$
$2^2=u_y^2-2\left(10\right)\left(0.4\right)$
$\Rightarrow u_y^2=12$ or $u_y=\sqrt{12}=2\sqrt{3}$
$\tan \theta =\frac{u_y}{u_x}=\frac{1}{\sqrt{3}}\Rightarrow \theta =30°$
Hence, the correct answer is option (c).