Angle of projection for the maximum horizontal range of a projectile is 45 degrees, if the point of

Q: Why do we convert the trajectory equation into a quadratic in $\tan\theta$?

Because for a fixed target $(x, y)$ and a fixed initial speed $u$, the only variable we can change is the launch angle $\theta$. The quadratic form allows us to use the discriminant to find the mathematical limits of reachability without needing calculus.

Q: What does the discriminant $D \ge 0$ physically mean?

It means that the target point $(x, y)$ is within the projectile's physical range. If $D < 0$, it means no real angle exists that can make the projectile hit that point with the given initial speed.

Q: Why does the maximum range occur exactly when $D = 0$?

$D=0$ represents the absolute boundary between reachable and unreachable points in space. At this boundary, there is exactly one unique angle to hit the target, which corresponds to the absolute maximum horizontal distance for that specific height.

Q: Can we solve this using calculus instead?

Yes! You can differentiate the trajectory equation implicitly with respect to $\theta$, set $dx/d\theta = 0$, and solve for $\theta$. However, the discriminant method is often much faster and more elegant algebraically.

Q: Why is the angle for maximum range on level ground $45^\circ$, but different here?

On level ground ($y=0$), our derived formula simplifies to $\sin\theta = u/\sqrt{2u^2} = 1/\sqrt{2}$, which gives exactly $45^\circ$. When the target is elevated or depressed, the optimal angle must shift to account for the extra vertical distance the projectile must travel against gravity.

Question

Angle of projection for the maximum horizontal range of a projectile is $45^\circ$, if the point of projection and the point of landing are in the same horizontal level. Determine the angle of projection for the maximum horizontal range of a projectile, if
(a) the point of landing is at a height $h$ above the point of projection.
(b) the point of landing is at a depth $h$ below the point of projection.

Sigma Percentile · Accepted Answer

### The Safety Parabola: Optimizing Projectile Range on Uneven Terrain

When you first learn about projectile motion, one of the most famous facts you memorize is that the maximum horizontal range on level ground is achieved at a launch angle of $45^\circ$. But what happens when the target is not on the same horizontal level? What if you are firing a cannon at a fortress on a hill, or throwing a stone into a deep valley?

The $45^\circ$ rule no longer applies. To find the optimal angle for an elevated or depressed target, we need a more powerful mathematical tool. Enter the **Discriminant Method**.

### The Trajectory Equation as a Quadratic

Let's start with the fundamental equation of trajectory, which describes the parabolic path of a projectile launched with speed $u$ at an angle $	heta$:

$$y = x 	an	heta - \frac{g x^2}{2 u^2 \cos^2	heta}$$

Our goal is to hit a specific target point $P(x, y)$. In this scenario, the coordinates $x$ and $y$ are fixed, and the initial speed $u$ is fixed. The only variable we can control is the launch angle $	heta$.

To make this equation easier to solve, we use a classic trigonometric identity: $\frac{1}{\cos^2	heta} = 1 + 	an^2	heta$. Substituting this into our trajectory equation gives:

$$y = x 	an	heta - \frac{g x^2}{2 u^2} (1 + 	an^2	heta)$$

By rearranging the terms, we can transform this into a beautiful quadratic equation in terms of $	an	heta$:

$$\frac{g x^2}{2 u^2} 	an^2	heta - x 	an	heta + \left( y + \frac{g x^2}{2 u^2} ight) = 0$$

### The Magic of the Discriminant

Now, think about what this quadratic equation means physically. For the projectile to actually reach the target point $(x, y)$, there must exist a real, physical launch angle $	heta$. In the language of algebra, this means our quadratic equation must have real roots.

For a quadratic equation to have real roots, its discriminant ($D = b^2 - 4ac$) must be greater than or equal to zero ($D \ge 0$). Let's set that up:

$$(-x)^2 - 4 \left( \frac{g x^2}{2 u^2} ight) \left( y + \frac{g x^2}{2 u^2} ight) \ge 0$$

Simplifying this inequality reveals a profound physical truth:

$$x^2 - \frac{2g x^2 y}{u^2} - \frac{g^2 x^4}{u^4} \ge 0$$

Dividing by $x^2$ (since the horizontal distance $x$ is positive) and rearranging the terms, we get an upper limit on how far the projectile can travel horizontally for a given height $y$:

$$x \le \frac{u}{g} \sqrt{u^2 - 2gy}$$

This boundary defines the **Envelope of Trajectories** or the **Safety Parabola**. Any point inside this envelope can be hit; any point outside is mathematically unreachable.

### Finding the Optimal Angle

The absolute maximum horizontal range $x_{max}$ occurs exactly at the boundary of this envelope, where the discriminant is exactly zero ($D = 0$).

When $D = 0$, a quadratic equation has a single repeated root given by $\frac{-b}{2a}$. Let's find the tangent of our optimal angle:

$$	an	heta = \frac{-(-x_{max})}{2 \left( \frac{g x_{max}^2}{2 u^2} ight)} = \frac{u^2}{g x_{max}}$$

Substituting our expression for $x_{max}$ into this formula gives:

$$	an	heta = \frac{u^2}{g \left( \frac{u}{g} \sqrt{u^2 - 2gy} ight)} = \frac{u}{\sqrt{u^2 - 2gy}}$$

To match standard multiple-choice options, we often need to convert this to sine. Using the right-triangle relationship $\sin	heta = \frac{	an	heta}{\sqrt{1 + 	an^2	heta}}$, we get a beautifully compact expression:

$$\sin	heta = \frac{u}{\sqrt{2u^2 - 2gy}}$$

### The Final Calculation

Now, we can easily answer both parts of the original question by simply plugging in the correct $y$ coordinate.

**(a) Target is at a height $h$ above the projection point:**
Substitute $y = h$ into our formula:

$$	heta = \sin^{-1}\left( \frac{u}{\sqrt{2(u^2 - gh)}} ight)$$

**(b) Target is at a depth $h$ below the projection point:**
Substitute $y = -h$ into our formula:

$$	heta = \sin^{-1}\left( \frac{u}{\sqrt{2(u^2 + gh)}} ight)$$

Notice how elegant this method is. We didn't need to use complex calculus or take derivatives. By simply demanding that the math reflects physical reality (real roots require a non-negative discriminant), the optimal angle naturally revealed itself!

Visualized Solution

$Visualizing the Target$

$The Equation of Trajectory$

$Forming the Quadratic Equation$

$The Condition for Reachability$

$Maximizing the Horizontal Range$

$Optimal Angle for Maximum Range$

$Converting to Sine$

$Final Substitution$

$The Envelope of Trajectories$

The Sigma Insight: Projectile Motion

The Safety Parabola

The Trajectory Equation as a Quadratic

The Magic of the Discriminant

Finding the Optimal Angle

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Target

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Equation of Trajectory

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forming the Quadratic Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Condition for Reachability

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Maximizing the Horizontal Range

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Optimal Angle for Maximum Range

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Converting to Sine

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Envelope of Trajectories