Animated Solution for Physics - Kinematics: A particle projected from the ground passes two points, which are at heights $h_1=12$ m and $h_2=18$ m above the ground and a distance $d=10$ m apart. What could be the minimum speed of projection? Acceleration due to gravity is $g=10$ m/s${}^2$.

A Parabolic Puzzle

Imagine you are an artillery commander, and you need to fire a projectile that must pass precisely through two specific windows in two different buildings. You know the heights of these windows, $h_1$ and $h_2$, and you know the straight-line distance $d$ between them. Your goal is to accomplish this feat using the absolute minimum possible projection speed from the ground.

At first glance, this seems like a nightmare of algebraic manipulation. If you try to plug the coordinates of the two points into the standard equation of trajectory, $y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$, you will quickly find yourself drowning in a sea of trigonometric identities and complex optimization conditions. While it is mathematically possible to find the minimum speed $u$ by forcing the discriminant of the resulting quadratic equation to be non-negative, there is a much more elegant, purely physical way to look at the problem.

The Virtual Incline

Instead of trying to analyze the entire parabolic path starting from the ground, let's shift our perspective. Let's focus exclusively on the segment of the trajectory that lies between the two points, $P_1$ and $P_2$.

Imagine drawing a straight line connecting $P_1$ and $P_2$. Now, treat this line as a virtual inclined plane. The particle, in order to pass through both points, must effectively travel a distance $d$ along this inclined plane.

What is the angle of this incline? Let's call it $\alpha$. From simple geometry, we can see that the vertical rise between the two points is $h_2-h_1$, and the hypotenuse is $d$. Therefore, the sine of the angle of inclination is given by:

Now, we ask a simpler question: What is the minimum speed $v_1$ required at point $P_1$ to just barely reach point $P_2$ on this incline?

We know from standard projectile motion theory that the maximum range $R_{max}$ up an inclined plane of angle $\alpha$ for a given initial velocity $v$ is:

To minimize the required velocity $v_1$ for a fixed distance $d$, we must assume that $d$ is exactly the maximum possible range for that velocity. Setting $R_{max}=d$ and rearranging for $v_1^2$, we get:

This is the absolute minimum squared speed the particle must possess as it passes through the first point $P_1$. Let's substitute our geometric expression for $\sin \alpha$ into this equation:

Distributing the $d$ inside the parenthesis yields a beautifully simplified expression:

Energy Conservation

We have found the minimum speed required at $P_1$, but the question asks for the minimum projection speed $u$ from the ground. How do we bridge this gap?

This is where the Law of Conservation of Mechanical Energy shines. The total mechanical energy of the projectile at the moment of launch must equal its total mechanical energy as it passes through $P_1$. Assuming the ground is our reference level for zero potential energy, we can write:

Dividing by the mass $m$ and multiplying by 2, we get a direct relationship between the squared speeds:

Now for the grand synthesis. We substitute our expression for $v_1^2$ into the energy equation:

Expanding the terms:

Notice how the $-g{h}_1$ and $+2g{h}_1$ perfectly combine to give $+g{h}_1$. The result is an incredibly elegant, symmetric master formula for the minimum projection speed:

This formula is a thing of beauty. It tells us that the minimum squared speed is simply the acceleration due to gravity multiplied by the sum of the two heights and the distance between them.

Let's execute the final calculation with the given values: $h_1=12$ m, $h_2=18$ m, $d=10$ m, and $g=10{\text{ m/s}}^2$.

Equation of Trajectory

While the inclined plane analogy is a brilliant shortcut, it is worth noting that this result can also be derived rigorously using the equation of trajectory. By forcing the trajectory to pass through $(x_1,h_1)$ and $(x_2,h_2)$, and applying the condition that the required projection angle $\theta$ must be real (discriminant $\ge 0$), you will eventually arrive at the exact same formula.

Furthermore, this formula is remarkably robust. It holds true even if the second point is lower than the first point ($h_2<h_1$). The symmetry of the formula $h_1+h_2$ guarantees that the order of the points does not matter, which perfectly aligns with the time-reversal symmetry of projectile motion.