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

Imagine you are standing on a flat ground and you throw a ball. The path it takes through the air is what we call its trajectory. In this problem, we are given the initial velocity of the projectile as a vector: $\vec{u}=\hat{i}+2\hat{j}$ m/s.

What does this vector tell us? It breaks down the initial push into two independent directions. The $\hat{i}$ component tells us the ball is moving horizontally at $1$ m/s. The $\hat{j}$ component tells us it is simultaneously moving upwards at $2$ m/s.

To find the equation of the trajectory, we need to find a relationship between the horizontal position $x$ and the vertical position $y$. The secret to projectile motion is that the horizontal and vertical motions are completely independent of each other, linked only by the time $t$ they have been in the air.

Let's write down the equations of motion for both directions. Horizontally, there is no acceleration (ignoring air resistance), so the distance covered is simply speed multiplied by time:

Vertically, gravity is pulling the projectile down, so we use the second equation of motion:

Now, let's substitute our known values into these equations. We know $u_x=1$, $u_y=2$, and $g=10$.

We have $x$ in terms of $t$, and $y$ in terms of $t$. But we want $y$ in terms of $x$. Since $t=x$, we can simply replace every $t$ in the $y$ equation with $x$. This is a powerful mathematical technique called parameter elimination.

Substituting $t=x$ into the vertical equation:

And there we have it! This is the equation of the trajectory. It's a quadratic equation, which confirms that the path is a parabola opening downwards.

The Standard Formula

If you prefer using formulas, you can also solve this using the standard equation of trajectory:

From our initial velocity vector, we can find the angle of projection $\theta$ and the initial speed $u$.

Notice that $u^2{\cos }^2\theta$ is simply the square of the horizontal velocity component, $u_x^2$, which is $1^2=1$.

Plugging these into the standard formula:

Both methods beautifully converge to the exact same result, proving the consistency of physics!