A cannon installed at the top of a hill can fire shells in all directions. There is an enemy bunker

Q: Why do we minimize the square of the distance instead of the distance itself?

Minimizing $s^2$ is mathematically equivalent to minimizing $s$ because distance is always positive. It saves us from dealing with messy square roots during differentiation or algebraic manipulation.

Q: Why did we group the terms into $A \cos\theta + B \sin\theta$?

This is a standard trigonometric identity trick. The expression $A \cos\theta + B \sin\theta$ can be rewritten as $\sqrt{A^2+B^2} \sin(\theta + \alpha)$, which clearly has a maximum value of $\sqrt{A^2+B^2}$ when the angle is such that $\tan\theta = B/A$.

Q: What if the shell hits the ground before time $T$?

The problem implicitly assumes that the shell explodes in the air before hitting the ground or the hill. If it hit the ground first, the trajectory equations would no longer be valid after the impact.

Q: Does the initial speed $u$ affect the optimal angle?

Interestingly, no! If you look at the final expression for $\theta$, the initial speed $u$ completely cancels out during the maximization step. The optimal angle depends only on the target's position, gravity, and the fuse time $T$.

Q: How does this relate to firing directly at the target?

If there were no gravity ($g=0$), the formula gives $\theta = \phi$, which means aiming directly at the target. The extra term $\frac{gT^2}{2d\cos\phi}$ is the exact angular correction needed to compensate for the shell falling under gravity during its flight.

Question

A cannon installed at the top of a hill can fire shells in all directions. There is an enemy bunker at an angle of elevation $\phi$ and a distance $d$ from the cannon. All the shells fired explode in air in time $T$ before they reach the bunker. At what angle to the horizontal, should a shell be fired with a speed $u$ to explode closest to the bunker? Acceleration due to gravity is $g$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an artillery commander tasked with hitting an enemy bunker located on a distant hill. The bunker is at a straight-line distance $d$ from your cannon, elevated at an angle $\phi$. Your shells are equipped with a time fuse set to explode exactly at time $T$. Your mission is to find the perfect firing angle $	heta$ so that the shell explodes as close to the bunker as possible.

Let's set up a coordinate system to translate this physical battlefield into mathematics. We place our cannon at the origin $(0,0)$. Using basic trigonometry, the coordinates of the enemy bunker $B$ are:
$$x_B = d \cos\phi$$
$$y_B = d \sin\phi$$

### The Master Equation

When the cannon fires a shell with an initial speed $u$ at an angle $	heta$, the shell follows a parabolic trajectory dictated by gravity. The position of the shell at any time $t$ is given by the standard kinematic equations:
$$x(t) = (u \cos	heta) t$$
$$y(t) = (u \sin	heta) t - \frac{1}{2} g t^2$$

Since the shell is timed to explode at $t = T$, the exact coordinates of the explosion point $E$ will be:
$$x_E = u T \cos	heta$$
$$y_E = u T \sin	heta - \frac{1}{2} g T^2$$

Our objective is to minimize the distance $s$ between the explosion point $E$ and the bunker $B$. To avoid dealing with messy square roots, we can equivalently minimize the square of the distance, $S = s^2$:
$$S = (x_E - x_B)^2 + (y_E - y_B)^2$$

### Expanding and Simplifying

Substituting our coordinates into the distance squared formula gives us a rather intimidating expression:
$$S = (u T \cos	heta - d \cos\phi)^2 + \left(u T \sin	heta - \frac{1}{2} g T^2 - d \sin\phiight)^2$$

Don't let the algebra scare you! Let's expand the squares carefully. When we do, we will encounter terms like $u^2 T^2 \cos^2	heta$ and $u^2 T^2 \sin^2	heta$. Thanks to the Pythagorean identity $\cos^2	heta + \sin^2	heta = 1$, these combine beautifully into a single constant term $u^2 T^2$.

In fact, if we group all the terms that do not depend on our variable firing angle $	heta$ into a single constant $C$, the expression simplifies to:
$$S = C - 2uT \left[ d \cos\phi \cos	heta + \left(d \sin\phi + \frac{1}{2} g T^2ight) \sin	heta ight]$$

### The Optimization Trick

To make the total distance squared $S$ as small as possible, we must **maximize** the term being subtracted. Let's call this term $f(	heta)$:
$$f(	heta) = A \cos	heta + B \sin	heta$$
where $A = d \cos\phi$ and $B = d \sin\phi + \frac{1}{2} g T^2$.

This is a classic optimization scenario in physics and mathematics! The maximum value of any expression in the form $A \cos	heta + B \sin	heta$ occurs when:
$$	an	heta = \frac{B}{A}$$

### Final Calculation

Now, we simply substitute our expressions for $A$ and $B$ back into this condition:
$$	an	heta = \frac{d \sin\phi + \frac{1}{2} g T^2}{d \cos\phi}$$

Dividing both terms in the numerator by the denominator $d \cos\phi$, we get:
$$	an	heta = 	an\phi + \frac{g T^2}{2d \cos\phi}$$

Taking the inverse tangent gives us our final, elegant result:
$$	heta = 	an^{-1}\left(	an\phi + \frac{g T^2}{2d \cos\phi}ight)$$

**Physical Insight:** Notice how the initial speed $u$ completely vanished from our final answer! Furthermore, if we were in a universe without gravity ($g=0$), the formula simplifies to $	heta = \phi$, meaning we would aim directly at the target. The additional term $\frac{g T^2}{2d \cos\phi}$ is the precise angular correction required to compensate for the shell falling under the influence of gravity during its flight time $T$.

Visualized Solution

$C oor d ina t e S y s t e m S e t u p$

$P r o j ec t i l e M o t i o n E q u a t i o n s$

$C oor d ina t es o f E x pl os i o n P o in t$

$D i s t an ce B e tw ee n E x pl os i o n an d B u nk er$

$E x p an d in g t h e D i s t an ce E x p r ess i o n$

$S im pl i f y in g t h e E x p r ess i o n$

$M inimi z in g t h e D i s t an ce$

$F ina l C a l c u l a t i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Projectile Motion

Analyzing the Setup

The Master Equation

Expanding and Simplifying

The Optimization Trick

Final Calculation

Similar Questions

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The trajectory of a projectile near the surface of the earth is given as $y = 2 x - 9 x^{2}$ . If it were launched at an angle $θ_{0}$ with speed $v_{0}$ , then (Take, $g = 10 ms^{- 2}$ )

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Coordinates of Explosion Point

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Distance Between Explosion and Bunker

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

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

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

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

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