Question

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

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

Answer 1

Answer: (C)

$v^2=u^2-2gh$ or $u^2=v^2+2gh$

or $u_x^2+u_y^2=v_x^2+v_y^2+2gh$

As $v_x=u_x;u_y^2=v_y^2+2gh$

or $u_y^2=(2{)}^2+2\times 10\times 0.4=12$

$u_y=\sqrt{12}=2\sqrt{3}m/s$

$u_x=v_x=6m/s$

tan $\theta =\frac{u_y}{u_x}=\frac{2\sqrt{3}}{6}=\frac{1}{\sqrt{3}}$

$\theta =30^{\circ }$