The maximum range of a shell fired from a gun is R = 22.5 m. This gun is mounted on a platform that

Q: Why doesn't the platform's velocity affect the time of flight?

The time of flight is determined entirely by the vertical component of the shell's velocity. Since the platform moves purely horizontally, it adds nothing to the vertical velocity, leaving the time of flight unchanged.

Q: Why do we use $R = u^2/g$ for the stationary gun?

For a projectile fired from the ground, the range is $u^2 \sin(2\theta)/g$. The maximum value of $\sin(2\theta)$ is 1 (when $\theta = 45^\circ$), making the maximum range exactly $u^2/g$.

Q: Can we just add the maximum ranges of the gun and the platform?

No! Range is not a linear function of velocity. You must add the velocity vectors first, and then calculate the new range based on the resultant velocity components.

Q: Why did we divide by $u^2$ in the quadratic equation?

Dividing by $u^2$ simplifies the equation by creating a dimensionless ratio $v/u$. This makes the algebra much cleaner and reveals that the optimal angle depends only on the ratio of the platform's speed to the muzzle velocity.

Q: What if the platform was moving in the opposite direction?

If the platform moved backwards, the net horizontal velocity would be $u \cos\theta - v$. The derivation would be identical, but the ratio $v/u$ in the quadratic equation would have a negative sign, leading to a different optimal angle.

Question

The maximum range of a shell fired from a gun is $R = 22.5 	ext{ m}$. This gun is mounted on a platform that can move horizontally with a constant speed $v = 15.0 	ext{ m/s}$. At what angle above the horizontal, must the gun be aimed to achieve maximum horizontal range? Neglect air resistance as well as height of the gun. Acceleration of free fall $g = 10.0 	ext{ m/s}^2$.

Sigma Percentile · Accepted Answer

Welcome to one of the most fascinating problems in kinematics! Imagine you are standing on a vast, open field, observing a military exercise. A gun is mounted on a heavy platform that is cruising horizontally at a constant speed. The challenge? To find the perfect angle to fire a shell so that it travels the absolute maximum distance across the ground.

This problem beautifully intertwines the concepts of relative velocity, projectile motion, and mathematical optimization. Let's break it down step by step and uncover the elegance hidden within the equations.

### The Moving Platform Conundrum

Before we deal with the moving platform, we must understand the gun itself. We are given a crucial piece of information: if the gun were completely stationary, its maximum range would be $22.5 	ext{ m}$.

From standard projectile motion, we know that the maximum range of a projectile fired from the ground is achieved at an angle of $45^\circ$, and this maximum range is given by the formula:

$$R = \frac{u^2}{g}$$

Here, $u$ is the muzzle velocity of the shell (its speed relative to the gun barrel), and $g$ is the acceleration due to gravity.

### Decoding the Muzzle Velocity

Let's use this information to find $u$. By substituting the given values $R = 22.5 	ext{ m}$ and $g = 10 	ext{ m/s}^2$ into our formula, we get:

$$u^2 = 22.5 	imes 10 = 225$$

Taking the square root, we find that the muzzle velocity is exactly $u = 15 	ext{ m/s}$.

Coincidentally, the problem states that the platform is also moving horizontally with a speed of $v = 15 	ext{ m/s}$. This symmetry ($u = v$) will make our mathematical journey incredibly satisfying later on.

### The Net Velocity Vector: A Galilean Dance

Now, let's set the platform in motion. The gun fires the shell at an angle $	heta$ relative to the horizontal platform.

According to Galilean relativity, the velocity of the shell relative to the ground is the vector sum of its velocity relative to the platform and the platform's velocity relative to the ground.

Let's break this down into components. The vertical motion is entirely unaffected by the platform's horizontal movement. Therefore, the vertical component of the shell's velocity is simply:

$$v_y = u \sin	heta$$

However, the horizontal motion gets a massive boost! The shell inherits the platform's forward speed. The net horizontal velocity becomes:

$$v_x = u \cos	heta + v$$

### The Master Equation for Range

The time the shell spends in the air—its time of flight $T$—is dictated solely by gravity and its initial vertical velocity. The platform's horizontal motion cannot keep the shell in the air any longer. Thus:

$$T = \frac{2v_y}{g} = \frac{2u \sin	heta}{g}$$

The horizontal range $R'$ on the ground is the product of the net horizontal velocity and the time of flight:

$$R' = v_x \cdot T = (u \cos	heta + v) \left(\frac{2u \sin	heta}{g}ight)$$

Let's expand this expression to see its structure:

$$R' = \frac{2u^2 \sin	heta \cos	heta}{g} + \frac{2uv \sin	heta}{g}$$

Using the double angle trigonometric identity $2 \sin	heta \cos	heta = \sin 2	heta$, we can rewrite the range as:

$$R' = \frac{u^2}{g} \sin 2	heta + \frac{2uv}{g} \sin	heta$$

### Calculus to the Rescue: Maximizing the Range

We have a function for the range $R'$ in terms of the firing angle $	heta$. To find the angle that maximizes this range, we must invoke the power of calculus. We take the derivative of $R'$ with respect to $	heta$ and set it to zero:

$$\frac{dR'}{d	heta} = \frac{u^2}{g} (2 \cos 2	heta) + \frac{2uv}{g} \cos	heta = 0$$

We can divide the entire equation by $2u/g$ to simplify it:

$$u \cos 2	heta + v \cos	heta = 0$$

To solve this, we need everything in terms of a single trigonometric function. We use another double angle identity, $\cos 2	heta = 2\cos^2	heta - 1$:

$$u (2\cos^2	heta - 1) + v \cos	heta = 0$$

Rearranging the terms, we arrive at a beautiful quadratic equation in terms of $\cos	heta$:

$$2u \cos^2	heta + v \cos	heta - u = 0$$

Dividing by $u$, we get:

$$2\cos^2	heta + \frac{v}{u}\cos	heta - 1 = 0$$

### The Final Strike: Solving the Quadratic

This is where the magic happens. Remember how we found that $u = 15 	ext{ m/s}$ and $v = 15 	ext{ m/s}$? Their ratio $v/u$ is exactly $1$. Substituting this into our quadratic equation yields:

$$2\cos^2	heta + \cos	heta - 1 = 0$$

This is a standard quadratic equation that can be easily factored:

$$(2\cos	heta - 1)(\cos	heta + 1) = 0$$

This gives us two possible mathematical solutions: $\cos	heta = 1/2$ or $\cos	heta = -1$.

Since the gun is fired above the horizontal, the angle $	heta$ must be acute (between $0^\circ$ and $90^\circ$). Therefore, $\cos	heta$ must be positive. We discard the negative root and are left with:

$$\cos	heta = \frac{1}{2}$$

Taking the inverse cosine, we find our ultimate answer:

**$$	heta = 60^\circ$$**

By aiming the gun at exactly $60^\circ$ relative to the moving platform, the shell will harness the perfect combination of vertical air time and horizontal momentum to achieve the absolute maximum range. Physics is truly elegant!

Visualized Solution

Understanding the Setup

Muzzle Velocity $u$

Calculating $u$

Velocity Components $v_{x}$ and $v_{y}$

Time of Flight $T$

Horizontal Range $R^{'}$

Expanding the Range $R^{'}$

Maximizing Range $\frac{d R ^{'}}{d θ} = 0$

Trigonometric Substitution

Quadratic Equation in $cos θ$

Substituting Values of $u$ and $v$

Solving for $θ$

The Sigma Insight: Projectile Motion

The Moving Platform Conundrum

Decoding the Muzzle Velocity

The Net Velocity Vector

The Master Equation for Range

Calculus to the Rescue

The Final Strike

Similar Questions

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, $g = 10 ms^{- 2}$ )

A shell is fired from a fixed artillery gun with an initial speed $u$ such that it hits the target on the ground at a distance $R$ from it. If $t_{1}$ and $t_{2}$ are the values of the time taken by it to hit the target in two possible ways, the product $t_{1} t_{2}$ is

Visualized Solution

Understanding the Setup

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

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

Velocity Components .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vx​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vy​

Time of Flight .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

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

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

Maximizing Range .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dθdR′​=0