In a battle field of ancient times, a soldier with a catapult stationed on the top of a very high cl

Q: Why is the drag acceleration $-k\vec{v}$ and not just $-k$?

The problem states the rate of reduction of speed per unit distance is $k$, so $dv/ds = -k$. Acceleration is $v(dv/ds)$, which gives $-kv$. Vectorially, it opposes motion, so $\vec{a}_{\text{drag}} = -k\vec{v}$.

Q: Why don't we need the height of the cliff?

The cliff is 'very high', meaning the stone falls for a long enough time that its horizontal velocity exponentially decays to zero. It reaches an asymptotic maximum horizontal distance, which is independent of any further vertical drop.

Q: What happens to the vertical velocity?

The vertical velocity initially decreases, becomes zero at the peak, and then increases downwards until it reaches a terminal velocity $g/k$. However, this doesn't affect the maximum horizontal range.

Q: Can we use standard projectile motion formulas here?

No, standard projectile formulas assume zero air resistance (constant horizontal velocity). Here, the horizontal velocity is continuously decreasing due to drag.

Q: What if the cliff wasn't very high?

If the cliff had a specific finite height, you would need to solve the vertical equation of motion to find the exact time $t$ it hits the ground, and then substitute that $t$ into the horizontal displacement equation.

Question

In a battle field of ancient times, a soldier with a catapult stationed on the top of a very high cliff notices camps of enemy close to the bottom of the cliff as shown in the figure. Stones can be launched from the catapult with a speed $u = 40 	ext{ m/s}$ at an angle $	heta = 60^\circ$ above the horizontal. If the air resistance reduces speed of the stone at a rate $k = 0.1 	ext{ (m/s)/m}$ and there is no wind, at what horizontal separation from the enemy camps should the soldier install the catapult to hit the enemy camps?

Sigma Percentile · Accepted Answer

### The Battlefield Setup

Imagine you are standing on a very high cliff in an ancient battlefield. You are tasked with operating a catapult to strike an enemy camp located far below.

When the stone is launched, it doesn't just follow a simple parabolic path under the influence of gravity. The air resistance is significant and constantly pushes back against its motion.

To hit the target precisely, we must understand how this drag force alters the stone's trajectory. Let's visualize the forces at play and decode the physics behind this ancient artillery.

### Decoding the Drag

The problem provides a very specific and interesting detail: the air resistance reduces the speed of the stone at a rate of $k = 0.1 	ext{ (m/s)/m}$.

This unit might look strange at first, but it tells us that the rate of change of speed with respect to distance is constant. Mathematically, we can write this as $\frac{dv}{ds} = -k$.

To find the acceleration due to this drag, we can use the chain rule of calculus. We know that acceleration $a = \frac{dv}{dt}$, which can be rewritten as $a = v \frac{dv}{ds}$.

Substituting our given rate, we get $a = v(-k) = -kv$.

Vectorially, since drag always opposes the direction of motion, the drag acceleration vector is **$\vec{a}_{	ext{drag}} = -k\vec{v}$**. This is a crucial realization that sets the stage for the rest of our solution!

### The Horizontal Journey

Our goal is to find the horizontal separation between the catapult and the enemy camp. Therefore, we need to break our motion down into components and focus entirely on the horizontal $x$-axis.

The horizontal component of the drag acceleration is simply $a_x = -k v_x$.

We can write this as a simple first-order differential equation: 
$$ \frac{dv_x}{dt} = -k v_x $$

Let's integrate this equation to find how the horizontal velocity changes over time. We separate the variables:
$$ \int_{u_x}^{v_x} \frac{dv_x}{v_x} = \int_0^t -k dt $$

Integrating both sides gives us $\ln\left(\frac{v_x}{u_x}ight) = -kt$.

Taking the exponential of both sides, we find the horizontal velocity as a function of time:
$$ v_x(t) = u_x e^{-kt} $$

This result shows that the horizontal velocity is **exponentially decaying**. The stone is constantly slowing down in the forward direction due to the relentless air resistance.

### The 'Very High Cliff' Insight

Now that we have the velocity, we can find the horizontal position. We write velocity as $v_x = \frac{dx}{dt}$ and integrate once more:
$$ \int_0^x dx = \int_0^t u_x e^{-kt} dt $$

Evaluating this integral gives us the horizontal displacement $x$ as a function of time:
$$ x(t) = \frac{u_x}{k} (1 - e^{-kt}) $$

Notice the term in the bracket. As time $t$ increases, the exponential term $e^{-kt}$ gets smaller and smaller.

Here is the catch! The problem describes the cliff as **"very high"**. This implies that the stone will be in the air for a very long time before it hits the ground.

Mathematically, we take the limit as time approaches infinity ($t 	o \infty$). In this limit, the exponential term vanishes completely ($e^{-kt} 	o 0$).

This means the horizontal position approaches a maximum limit. The stone cannot travel further than this distance, no matter how high the cliff is!
$$ x_{	ext{max}} = \frac{u_x}{k} = \frac{u \cos	heta}{k} $$

### The Final Strike

We have our master equation for the maximum horizontal range. Now, it's time for the final calculation.

Let's substitute the given values into our equation. The initial speed is $u = 40 	ext{ m/s}$, the launch angle is $	heta = 60^\circ$, and the drag constant is $k = 0.1 	ext{ s}^{-1}$.

$$ x_{	ext{max}} = \frac{40 \cos 60^\circ}{0.1} $$

Since $\cos 60^\circ = 0.5$, we have:
$$ x_{	ext{max}} = \frac{40 	imes 0.5}{0.1} = \frac{20}{0.1} = 200 	ext{ m} $$

The maximum horizontal separation is exactly **$200 	ext{ m}$**. To ensure a direct hit on the enemy camp, the soldier must install the catapult exactly 200 meters horizontally away from the target.

Visualized Solution

\text{Analyzing the Motion}

\text{Formulating Air Resistance}

\text{Horizontal Equation of Motion}

\text{Solving for Horizontal Velocity}

\text{Horizontal Displacement}

\text{Applying the 'Very High Cliff' Condition}

\text{What if the cliff was short?}

The Sigma Insight: Projectile Motion

The Battlefield Setup

Decoding the Drag

The Horizontal Journey

The 'Very High Cliff' Insight

The Final Strike

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