Animated Solution for Physics - Kinematics: The trajectory of a projectile in a vertical plane is $y=ax-b{x}^2$, where $a,b$ are constants, and $x$ and $y$ are respectively, the horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is ......... and the angle of projection from the horizontal is ......... .

Have you ever watched a basketball arc through the air and wondered about the hidden mathematics guiding its journey? In physics, we often describe projectile motion using time as a parameter. But what if we eliminate time and look purely at the spatial path—the trajectory? This problem gives us exactly that: the equation of a trajectory in a vertical plane, $y=ax-b{x}^2$. It looks like a simple quadratic equation, but hidden within its coefficients $a$ and $b$ are the secrets of the projectile's maximum height and its initial launch angle. Let's embark on a mathematical journey to decode this parabola!

The equation $y=ax-b{x}^2$ is the mathematical signature of a downward-opening parabola. Here, $x$ represents the horizontal distance traveled, and $y$ is the vertical height. The fact that there is a negative sign in front of the $x^2$ term is crucial—it tells us that the projectile goes up, reaches a peak, and then comes back down, exactly as gravity dictates.

Our mission is twofold: find the maximum height attained by the projectile, and determine the angle at which it was initially launched. To do this, we don't need complex kinematic equations involving time; we just need the power of differential calculus.

Finding the Peak

Think about the highest point of a projectile's flight. What is physically unique about this point? It's the moment the projectile stops rising and is about to start falling. For that infinitesimally small instant, its vertical velocity is zero.

Geometrically, if you were to draw a tangent line to the parabolic curve at this peak, that line would be perfectly horizontal. A horizontal line has a slope of zero. In the language of calculus, the slope of the curve is given by the derivative of $y$ with respect to $x$, denoted as $\frac{dy}{dx}$.

So, our condition for maximum height is beautifully simple:

Let's apply this to our trajectory equation. Differentiating $y=ax-b{x}^2$ with respect to $x$, we get:

To find the exact horizontal position where the peak occurs, we set this derivative to zero:

This tells us where the maximum height happens horizontally. But we want to know what that maximum height is.

The Maximum Height

To find the actual maximum height, which we'll call $H$, we simply take this $x$-coordinate and plug it back into our original trajectory equation. We are essentially asking the equation, "What is the value of $y$ when $x$ is $\frac{a}{2b}$?"

Let's substitute:

Now, we carefully expand the squared term:

Notice how one $b$ in the numerator cancels with one $b$ in the denominator of the second term:

To subtract these fractions, we need a common denominator, which is $4b$. We can rewrite the first term:

Subtracting the numerators gives us our first grand result:

Now for the second part of our mission: the angle of projection. Imagine the exact moment the projectile leaves the origin $(0,0)$. The angle its initial velocity vector makes with the horizontal ground is the angle of projection, let's denote it as ${\theta }_0$.

If you look at the trajectory curve, the direction of the initial velocity is exactly the direction of the tangent line at the starting point. Therefore, the tangent of the launch angle, $\tan {\theta }_0$, is equal to the slope of the curve at $x=0$.

We already have our slope formula:

We evaluate this derivative at the launch point, $x=0$:

This means the initial slope is simply $a$. Therefore:

To find the angle itself, we take the inverse tangent:

And there we have it! By simply looking at the coefficients of the parabolic trajectory, we used calculus to elegantly extract both the maximum height and the initial launch angle. It's a powerful reminder of how deeply math and physics are intertwined.