Question

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

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${45}^{\circ }$
• D. ${37}^{\circ }$

Answer 1

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

$v=v_x\hat{i}+v_y\hat{j}=6\hat{i}+2\hat{j}$
We know, in case of projectile motion, horizontal component of the velocity remains constant.
$\therefore v_x=u_x=6$
For $y$- component, using third equation of motion
$v_y^2=u_y^2-2\left(10\right)\left(0.4\right)\Rightarrow u_y^2=12u_y=\sqrt{12}$
$=2\sqrt{3}$
The angle of projection $\theta$ is,
$\tan \theta =\frac{u_y}{u_x}\tan \theta =\frac{2\sqrt{3}}{6}\theta ={\tan }^{-1}\frac{1}{\sqrt{3}}\Rightarrow \theta ={30}^{\circ }$