Question

At a height of $0.4m$ from the ground the velocity of a projectile in vector form is $\vec v = (6\hat i + 2\hat j)m/s$. The angle of projection is $(g = 10m/{s^2})$
A) ${45^\circ}$
B) ${60^\circ}$
C) ${30^\circ}$
D) ${\tan ^{ - 1}}(3/4)$

Answer 1

Hint: In these kinds of situations the first thing important is to notice the given problem statement. Especially this is a question regarding little manipulation. First, we just have to sort what we got and then we just have to interpret the missing quantities for the equation. In this question, we will get the answer simply by putting the values in the third equation of motion.

Complete step by step answer:
Given,
The height of the projection is $0.4m$
Velocity vector of the projection is $\vec v = (6\hat i + 2\hat j)m/s$
Firstly we have to use the second equation of motion in order to obtain the missing quantity,
${v_y}^2 = {u_y}^2 - 2gh$
Hence to obtain the value we have to put the values in the equation,
\[
  {v_y}^2 = {u_y}^2 - 2gh \\
\Rightarrow {u_y} = \sqrt {{v_y}^2 + 2gh} \\
\Rightarrow {u_y} = \sqrt {{2^2} + 2*10*0.4} \\
\Rightarrow {u_y} = \sqrt {4 + 8} \\
\Rightarrow {u_y} = 2\sqrt 3 m/s \\
 \]
Now we have to find the tangent in order to obtain the angle,
$
  \tan \theta = \dfrac{{{u_y}}}{{{u_x}}} \\
\Rightarrow \tan \theta = \dfrac{{2\sqrt 3 }}{3} \\
\Rightarrow \tan \theta = \dfrac{1}{{\sqrt 3 }} \\
\Rightarrow \tan \theta = {30^\circ} \\
 $
Hence option (C) is correct.

Note:
The most important thing here is to notice the given values, then we can easily derivate the missing quantities. Apart from that, it is necessary to observe what process is straight and easy to find the rest of the values, and then we can execute the formulas. In this type of question interpreting the correct formula to be used is necessary unless it will get twisted.