A projectile of mass 200 g is launched in a viscous medium at an angle 60 degrees with the horizonta

Q: Why don't we use the standard range formula $R = u^2 \sin(2\theta)/g$?

That formula assumes no air resistance! In a viscous medium, there is a continuous horizontal force slowing the projectile down, so the standard kinematic equations for constant acceleration do not apply.

Q: Why is gravity not included in the horizontal equation?

Gravity acts strictly downwards (in the vertical $y$-direction). Since horizontal and vertical motions are independent, gravity does not affect the horizontal force balance.

Q: How did we get $v_{0x} = 135$ m/s?

The initial velocity is 270 m/s at an angle of $60^\circ$. The horizontal component is $v_0 \cos(60^\circ) = 270 \times 0.5 = 135$ m/s.

Q: What does the term $e^{-t/2}$ represent physically?

It represents the exponential decay of the horizontal velocity. Because the drag force is proportional to velocity, the velocity decreases exponentially over time rather than linearly.

Q: Why did we convert 200 g to 0.2 kg?

Always use standard SI units! The drag coefficient $c$ is given in kg/s, so the mass must be in kg to ensure dimensional consistency.

Question

A projectile of mass 200 g is launched in a viscous medium at an angle $60^\circ$ with the horizontal, with an initial velocity of 270 m/s. It experiences a viscous drag force $\vec{F} = -c\vec{v}$ where the drag coefficient $c = 0.1$ kg/s and $\vec{v}$ is the instantaneous velocity of the projectile. The projectile hits a vertical wall after 2 s. Taking $e = 2.7$, the horizontal distance of the wall from the point of projection (in m) is ____

Sigma Percentile · Accepted Answer

Have you ever wondered why a real baseball doesn't travel as far as the physics equations predict? The culprit is air resistance, or **viscous drag**. In standard projectile motion problems, we happily ignore air resistance, leading to beautiful, symmetric parabolic trajectories. But in this problem, we are thrown into a viscous medium where the drag force is proportional to the velocity: $\vec{F} = -c\vec{v}$.

This completely changes the game! Let's break down how to tackle this realistic scenario step-by-step.

### Analyzing the Setup

We are given a projectile of mass $m = 200 	ext{ g} = 0.2 	ext{ kg}$ launched at an angle of $60^\circ$ with an initial velocity $v_0 = 270 	ext{ m/s}$. It travels through a viscous medium and hits a vertical wall after $t = 2 	ext{ s}$. Our goal is to find the horizontal distance $S_x$ to the wall.

The most crucial realization here is that **horizontal and vertical motions are independent**. Gravity acts downwards, affecting only the vertical motion. In the horizontal direction, the *only* force acting on the projectile is the horizontal component of the viscous drag.

### The Master Equation

Let's apply Newton's Second Law in the horizontal ($x$) direction. The net horizontal force is simply the horizontal drag force:

$$ F_x = -c v_x $$

Using $F_x = m a_x = m \frac{dv_x}{dt}$, we can set up our differential equation:

$$ m \frac{dv_x}{dt} = -c v_x $$

This is a classic first-order separable differential equation. Let's rearrange it to group the velocity terms on one side and the time terms on the other:

$$ \frac{dv_x}{v_x} = -\frac{c}{m} dt $$

### Solving for Velocity

Now, we integrate both sides. At $t = 0$, the initial horizontal velocity is $v_{0x}$. At time $t$, the velocity is $v_x$.

$$ \int_{v_{0x}}^{v_x} \frac{dv_x}{v_x} = \int_0^t -\frac{c}{m} dt $$

The integral of $1/v_x$ is the natural logarithm $\ln(v_x)$. Evaluating the limits gives:

$$ \ln\left(\frac{v_x}{v_{0x}}ight) = -\frac{c}{m} t $$

To isolate $v_x$, we take the exponential of both sides:

$$ v_x = v_{0x} e^{-\frac{c}{m} t} $$

This equation tells us a profound physical truth: in a viscous medium, the horizontal velocity doesn't stay constant; it **decays exponentially** over time!

### Finding the Displacement

We have the velocity as a function of time, but we need the horizontal displacement $S_x$. Since $v_x = \frac{dx}{dt}$, we can find the displacement by integrating the velocity function from $t = 0$ to $t = 2 	ext{ s}$.

$$ \int_0^{S_x} dx = \int_0^t v_{0x} e^{-\frac{c}{m} t} dt $$

Integrating the exponential function requires dividing by the coefficient of $t$:

$$ S_x = v_{0x} \left[ \frac{e^{-\frac{c}{m} t}}{-\frac{c}{m}} ight]_0^t $$

Applying the limits (and remembering that $e^0 = 1$), we get our master formula for horizontal displacement:

$$ S_x = \frac{m v_{0x}}{c} \left( 1 - e^{-\frac{c}{m} t} ight) $$

### Final Calculation

Now for the grand finale! Let's substitute our given values into the formula. First, let's calculate the constants:
- Mass $m = 0.2 	ext{ kg}$
- Drag coefficient $c = 0.1 	ext{ kg/s}$
- The ratio $\frac{c}{m} = \frac{0.1}{0.2} = 0.5 	ext{ s}^{-1}$
- Initial horizontal velocity $v_{0x} = v_0 \cos(60^\circ) = 270 	imes 0.5 = 135 	ext{ m/s}$

Substitute these into our displacement equation for $t = 2 	ext{ s}$:

$$ S_x = \frac{135}{0.5} \left( 1 - e^{-0.5 	imes 2} ight) $$

$$ S_x = 270 \left( 1 - e^{-1} ight) $$

$$ S_x = 270 \left( 1 - \frac{1}{e} ight) $$

The problem kindly provides the approximation $e = 2.7$. Let's plug that in:

$$ S_x = 270 \left( 1 - \frac{1}{2.7} ight) $$

$$ S_x = 270 \left( \frac{2.7 - 1}{2.7} ight) = 270 \left( \frac{1.7}{2.7} ight) $$

Notice how beautifully the numbers cancel out!

$$ S_x = 100 	imes 1.7 = \mathbf{170 	ext{ m}} $$

The projectile hits the wall at a horizontal distance of exactly 170 meters. By carefully setting up our differential equations and trusting the math, we conquered the viscous drag!

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Horizontal Equation of Motion

Setting up the Integral

Velocity as a Function of Time

Position as a Function of Time

Integrating for Horizontal Distance

Substituting the Given Values

Final Calculation

The Way Forward

The Sigma Insight: Newton's Laws of Motion

Analyzing the Setup

The Master Equation

Solving for Velocity

Finding the Displacement

Final Calculation

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Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Horizontal Equation of Motion

Setting up the Integral

Velocity as a Function of Time

Position as a Function of Time

Integrating for Horizontal Distance

Substituting the Given Values

Final Calculation

The Way Forward

The Sigma Insight: Newton's Laws of Motion

Analyzing the Setup

The Master Equation

Solving for Velocity

Finding the Displacement

Final Calculation

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