Animated Solution for Physics - Kinematics: The trajectory of a projectile near the surface of the earth is given as $y=2x-9x^2$. If it were launched at an angle ${\theta }_0$ with speed $v_0$, then (Take, $g=10{\text{ ms}}^{-2}$)

Imagine a ball thrown into the air. The path it traces is a beautiful parabola, dictated by the initial velocity and the relentless downward pull of gravity. In physics, we often describe this motion using time $t$ as a parameter. However, there is a more direct way to describe the path: the equation of trajectory. This equation relates the vertical position $y$ directly to the horizontal position $x$, completely eliminating time from the picture.

The standard equation of trajectory for a projectile launched from the origin with an initial speed $v_0$ at an angle ${\theta }_0$ is given by:

This equation is incredibly powerful because it is a unique polynomial representation of the path. If you are given any quadratic equation describing a projectile's path, you can simply compare it term-by-term with this standard equation to extract all the physical parameters of the launch.

The Linear Term

In our problem, we are given the specific trajectory equation:

Let's start by comparing the linear terms (the terms with just $x$). In the standard equation, the coefficient of $x$ is $\tan {\theta }_0$. In our given equation, the coefficient of $x$ is $2$. Therefore, we can immediately state:

This tells us the initial slope of the launch. But to find the initial speed $v_0$, we will need $\cos {\theta }_0$. How do we get that without calculating messy inverse tangents?

Whenever you know one trigonometric ratio, you can find all others by sketching a simple right-angled triangle. Since $\tan {\theta }_0=\frac{\text{Opposite}}{\text{Adjacent}}=\frac{2}{1}$, imagine a triangle where the side opposite to ${\theta }_0$ is $2$ and the adjacent side is $1$.

Using the Pythagorean theorem, the hypotenuse is:

Now, we can easily read off the cosine of the angle:

This elegant trick saves time and prevents calculation errors.

The Quadratic Term

Now, let's look at the quadratic terms (the terms with $x^2$). In the standard equation, the coefficient is $-\frac{g}{2v_0^2{\cos }^2{\theta }_0}$. In our given equation, it is $-9$. Equating the magnitudes, we get:

We are given $g=10{\text{ m/s}}^2$, and we just found that $\cos {\theta }_0=\frac{1}{\sqrt{5}}$, which means ${\cos }^2{\theta }_0=\frac{1}{5}$. Let's substitute these values into our equation:

Now it's just a matter of careful algebra. The $\frac{1}{5}$ in the denominator flips up to multiply the numerator:

Rearranging to solve for $v_0^2$:

Taking the square root of both sides, we find the initial speed:

And from our earlier triangle work, we know the angle is:

By systematically comparing coefficients and using basic trigonometry, we've completely decoded the projectile's launch parameters!