A particle just clears a wall of height b at a distance a and strikes the ground at a distance c fro

Q: Wait, why can't we just use energy conservation to find the velocity?

That is a great thought, but energy conservation only tells us about the speed at a certain height, not the direction of motion. In projectile motion, we need to account for both the horizontal and vertical components of velocity simultaneously. The trajectory equation is a much more direct path because it links the spatial coordinates $(x, y)$ directly to the launch angle $\theta$ and the range $R$.

Q: How did you get that trajectory equation $y = x \tan \theta (1 - \frac{x}{R})$?

It is derived from the standard equation $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$. By substituting the expression for the range $R = \frac{u^2 \sin 2\theta}{g}$, we can replace the term $\frac{g}{2u^2 \cos^2 \theta}$ with $\frac{\tan \theta}{R}$. It is a standard, highly useful shortcut in JEE physics that saves you from dealing with $u$ and $g$ entirely!

Q: Why is the range $R$ equal to $c$?

The problem states that the particle strikes the ground at a distance $c$ from the point of projection. By definition, the horizontal distance covered by a projectile when it returns to the same vertical level as the launch point is the horizontal range. Since the particle is launched from the ground and lands on the ground, the distance $c$ is exactly the range $R$.

Question

A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection. The angle of projection is

Options:

(A) $	an^{-1} \frac{bc}{a(c-a)}$

(B) $	an^{-1} \frac{bc}{a}$

(C) $	an^{-1} \frac{b}{ac}$

(D) 45°

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a flat, open field, holding a ball. You throw it into the air, and it traces a perfect, symmetric arc against the sky. This is the classic dance of projectile motion—a beautiful interplay between the constant, downward pull of gravity and the steady, forward momentum of inertia.

In this problem, we are not just looking at a simple arc; we are looking at a specific constraint: a wall of height $b$ at distance $a$ that our projectile must just barely clear. This adds a layer of precision to our physics. We are not just throwing; we are aiming.

### Defining the Coordinates

Let us define our world. We place the point of projection at the origin, $(0, 0)$. The particle is launched with an initial velocity $u$ at an angle $	heta$ to the horizontal.

As it travels, it follows a parabolic path. The wall, standing at a horizontal distance $x = a$, has a height $y = b$. The phrase "just clears" is our mathematical anchor; it tells us that the point $(a, b)$ lies exactly on the trajectory of the particle.

### Incorporating the Range

The problem gives us one more vital piece of information: the particle strikes the ground at a distance $c$ from the projection point. In the language of kinematics, this total horizontal distance is the Range, denoted by $R$.

Thus, we know $R = c$. Now we have all the pieces: the launch angle $	heta$, the point $(a, b)$ on the path, and the range $R = c$.

### The Elegant Tool

If we use the standard trajectory equation, $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$, we hit a wall of our own—we do not know the initial velocity $u$ or the acceleration due to gravity $g$. We need a more sophisticated tool.

Let us use the "Range-form" of the trajectory equation:

$$y = x 	an 	heta \left( 1 - \frac{x}{R} ight)$$

This equation is a masterpiece of algebraic efficiency. It relates the vertical height $y$ and horizontal distance $x$ to the launch angle $	heta$ and the range $R$, completely bypassing the need for $u$ and $g$.

### The Algebraic Victory

Now, we substitute our known values. We set $x = a$, $y = b$, and $R = c$. Our equation becomes:

$$b = a 	an 	heta \left( 1 - \frac{a}{c} ight)$$

Let us focus on the term inside the parentheses: $\left( 1 - \frac{a}{c} ight)$. By finding a common denominator, this becomes $\frac{c - a}{c}$.

Now, our equation is:

$$b = a 	an 	heta \left( \frac{c - a}{c} ight)$$

Our mission is to isolate $	an 	heta$. We multiply both sides by $c$ to get $bc = a 	an 	heta (c - a)$. Finally, we divide by $a(c - a)$ to find:

$$	an 	heta = \frac{bc}{a(c - a)}$$

Taking the inverse tangent of both sides, we arrive at the final expression for the angle of projection:

$$	heta = 	an^{-1} \left( \frac{bc}{a(c - a)} ight)$$

This is the angle of projection. It is a result that feels inevitable once you see the underlying structure. You have successfully navigated the constraints, used the right tool, and arrived at the solution.

Animated Solution for Mathematics - Trigonometry: A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection. The angle of projection is

Select Answer:

Visualized Solution

Visualizing the Projectile's Path

Introducing the Wall

Identifying the Range

The Standard Trajectory Equation

Trajectory in Terms of Range

Substituting the Known Values

Simplifying the Bracketed Term

Isolating $tan θ$

Finding the Angle of Projection $θ$

The Sigma Insight: Heights and Distances

Analyzing the Setup

Defining the Coordinates

Incorporating the Range

The Elegant Tool

The Algebraic Victory

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