Question

A projectile is thrown at some angle with the ground. If at a height of 0.4 m from the ground the velocity vector is equal to $\overrightarrow{v}=(6\hat{i}+2\hat{j})$ m/s then angle of projection from the ground will be (Take $g=10m/s^2$)

• A. ${30}^0$
• B. ${45}^0$
• C. ${53}^0$
• D. ${60}^0$

Answer 1

Answer: (A)

${30}^0$

Given,

$h=0.4m$

$g=10m/s^2$

$\overrightarrow{v}=6\hat{i}+2\hat{j}$ $m/s$

The horizontal component of velocity,

$v_x=6m/s$

The vertical component of velocity

$v_y=2m/s$

As, we know that the horizontal component remains constant,

$v_x=u_x=6m/s$

Vertical component,

$v{{}_y}^2=u{{}_y}^2+2a_{y}h$

$2^2=u{{}_y}^2-2\times 10\times 0.4$

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

Now, the angle of projecton is, $tan\theta =\frac{u_y}{u_x}$

$\theta =ta{n}^{-1}\left(\frac{2\sqrt{3}}{6}\right)$

$\theta ={30}^0$

The correct option is A.