Question

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

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

Answer 1

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

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

$\Rightarrow u_y=\sqrt{v_y^2+2gh}=\sqrt{2^2+2\left(10\right)\left(0.4\right)}$

$=\sqrt{4+8}=2\sqrt{3}m/s:$

$tan\theta =\frac{u_y}{u_x}=\frac{2\sqrt{3}}{6}=\frac{1}{\sqrt{3}}\Rightarrow ={30}^{\circ }$