Question

The position coordinates of a projectile from the ground on a certain planet (with no atmosphere) are given by $y=\left(4t-2t^2\right)m$ and $x=\left(3t\right)m$, where t is in second and point of projection is taken as the origin. The angle of projection of projectile with vertical is
$$\begin{gathered} \text{A}{\text{. 30}}^{\circ } \\ \text{B}{\text{. 37}}^{\circ } \\ \text{C}{\text{. 45}}^{\circ } \\ \text{D}{\text{. 60}}^{\circ } \end{gathered}$$

Answer 1

Hint: We are given the position coordinates of the projectile from which we can calculate the velocities of the particle along x and y-direction. Then from the projectile diagram, we can relate the angle with vertical to the velocities along the x and y-direction. Solving the obtained expression, we can find out the value of the angle.

Complete step-by-step solution:
We are given the expressions for the position coordinates of a projectile from the ground on
a certain point. The expression for the x-coordinate is given as
$x=\left(3t\right)m$
The expression for the y-coordinate is given as
$y=\left(4t-2t^2\right)m$
As we know the velocity of a particle is equal to the time derivative of the position of the particle. Therefore, we can find out the velocity of the given projectile along the x and y-direction. The velocity along the x-direction can be calculated in the following way.
$v_x={\displaystyle \frac{dx}{dt}}={\displaystyle \frac{d}{dt}}\left(3t\right)=3m/s$
Similarly, the velocity along the y-direction is given as
$v_y={\displaystyle \frac{dy}{dt}}={\displaystyle \frac{d}{dt}}\left(4t-2t^2\right)=4-4t$
We are asked to calculate the angle at origin and at the origin, t = 0. Therefore, we have the following values of velocities at t = 0.
$$\begin{gathered} \Rightarrow v_x=3m/s \\ \Rightarrow v_y=4-4\times 0=4m/s \end{gathered}$$

In the diagram, $\theta$ represents the angle made by the projectile with the vertical direction. Now according to the diagram for the projectile, we can write the following expression.
$$\begin{gathered} \Rightarrow \tan \theta ={\displaystyle \frac{v_x}{v_y}}={\displaystyle \frac{3}{4}} \\ \Rightarrow \theta ={\tan }^{-1}{\displaystyle \frac{3}{4}}={37}^{\circ } \end{gathered}$$
This is the required value of angle. Hence, the correct answer is option B.

Note: It should be noted that the absence of atmosphere means that there is no air friction between the projectile and the medium in which it is projected. If there is some friction present then it will hinder the motion of the projectile and it will end up covering a shorter distance.