A projectile is given an initial velocity of $$(\hat{i} + 2\hat{j})$$ m s$$^{-1}$$, where $$\hat{i}$$ is along the ground and $$\hat{j}$$ is along the vertical upward. If $$g = 10$$ m s$$^{-2}$$, the equation of its trajectory is :

We are told that the projectile is launched from the origin with the velocity vector

$$\vec u \;=\; 1\,\hat i \;+\; 2\,\hat j \quad\text{m s}^{-1}.$$

Here $$\hat i$$ is the horizontal (ground) unit vector and $$\hat j$$ is the vertical upward unit vector. Hence the horizontal and vertical components of the initial velocity are

$$u_x = 1\ \text{m s}^{-1}, \qquad u_y = 2\ \text{m s}^{-1}.$$

For a projectile that starts from the origin, the general equation of the trajectory (the path in the $$x\!-\!y$$ plane) is obtained from kinematics. First we write the two independent component equations of motion:

Horizontal motion (no acceleration horizontally):

$$x = u_x\,t.$$

Vertical motion (constant downward acceleration $$g$$):

$$y = u_y\,t - \dfrac{1}{2}g\,t^2.$$

To eliminate the time $$t$$, we first solve the horizontal equation for $$t$$:

$$t = \dfrac{x}{u_x}.$$

Now we substitute this value of $$t$$ into the vertical equation:

$$$ \begin{aligned} y &= u_y\left(\dfrac{x}{u_x}\right) \;-\; \dfrac{1}{2}g\left(\dfrac{x}{u_x}\right)^2 \\[6pt] &= \dfrac{u_y}{u_x}\,x \;-\; \dfrac{g}{2u_x^{\,2}}\,x^2. \end{aligned} $$$

This is the standard textbook form

$$y = x\,\tan\theta \;-\; \dfrac{g}{2u^{\,2}\cos^2\theta}\,x^2,$$

but expressed directly through the components $$u_x, u_y$$ it reads

$$y = \dfrac{u_y}{u_x}\,x \;-\; \dfrac{g}{2u_x^{\,2}}\,x^2.$$

Now we plug in the numerical values $$u_x = 1\ \text{m s}^{-1},\; u_y = 2\ \text{m s}^{-1},\; g = 10\ \text{m s}^{-2}:$$

$$$ \begin{aligned} y &= \dfrac{2}{1}\,x \;-\; \dfrac{10}{2\,(1)^2}\,x^2 \\[6pt] &= 2x \;-\; \dfrac{10}{2}\,x^2 \\[6pt] &= 2x \;-\; 5x^2. \end{aligned} $$$

Thus the explicit equation of the trajectory is

$$y = 2x - 5x^2.$$

Comparing this with the given options, we see it matches option D.

Hence, the correct answer is Option D.