A marble is projected in a viscous fluid, with an initial speed u at an angle theta above the horizo

Q: Why is the horizontal acceleration only $-k v_x$ and not $-k v$?

The drag force is a vector $\vec{a}_D = -k\vec{v}$. When we resolve this vector into components, the horizontal component is $-k v_x$ and the vertical component is $-k v_y$. Gravity only has a vertical component.

Q: How do we integrate $\frac{dv_y}{k v_y + g}$?

You can use a simple substitution. Let $u = k v_y + g$, then $du = k dv_y$. The integral becomes $\frac{1}{k} \int \frac{du}{u}$, which evaluates to $\frac{1}{k} \ln|u|$.

Q: Why does the horizontal velocity eventually become zero?

Because there is no forward force to keep it moving, and the drag force continuously removes kinetic energy. Mathematically, the $e^{-kt}$ term decays to zero as time goes on.

Q: What exactly is terminal velocity?

Terminal velocity is the constant speed an object reaches when the drag force exactly balances the driving force (in this case, gravity). At this point, the net acceleration is zero, so the velocity stops changing.

Q: Can the marble ever move backwards horizontally?

No. The drag force only opposes motion; it cannot reverse it. As $v_x$ approaches zero, the drag force in the horizontal direction also approaches zero.

Question

A marble is projected in a viscous fluid, with an initial speed $u$ at an angle $	heta$ above the horizontal. Drag force of the fluid results in an acceleration $\vec{a}_D = -k\vec{v}$ in addition to that of gravity, where $k$ is a positive constant and $\vec{v}$ is the velocity of marble. Determine position coordinates and $x$ (horizontal) and $y$ (vertical) components of velocity of the marble as functions of time $t$. What is the terminal velocity?

Sigma Percentile · Accepted Answer

### The Setup: A Battle of Forces

Imagine shooting a marble into a pool of thick honey. It doesn't just fly freely like a projectile in a vacuum; it has to fight its way through.

The fluid exerts a drag force that constantly saps its speed. We are given that this drag causes a retardation of:

$$ \vec{a}_D = -k\vec{v} $$

And of course, gravity is always there, pulling it down with an acceleration of $\vec{g}$.

### Divide and Conquer: Splitting the Motion

To solve any 2D kinematics problem, our best strategy is to split the motion into independent horizontal and vertical components.

In the horizontal direction, there is no gravity. The only force acting on the marble is the horizontal component of the drag. Thus, the horizontal acceleration is simply:

$$ a_x = -k v_x $$

In the vertical direction, it's a double whammy. The marble faces both the vertical component of the drag and the relentless pull of gravity. So, the vertical acceleration becomes:

$$ a_y = -k v_y - g $$

### Taming the Horizontal Motion

Let's tackle the horizontal motion first. We know that acceleration is the derivative of velocity, so we can write:

$$ \frac{dv_x}{dt} = -k v_x $$

This is a classic first-order separable differential equation. By rearranging the terms, we get:

$$ \frac{dv_x}{v_x} = -k dt $$

Integrating both sides from the initial horizontal velocity $u_x = u\cos	heta$ to some velocity $v_x$ at time $t$, we find:

$$ \ln\left(\frac{v_x}{u_x}ight) = -kt $$

Exponentiating both sides gives us the horizontal velocity as a function of time:

$$ v_x = u_x e^{-kt} $$

Notice the beauty of this result! The horizontal velocity decays exponentially. It will never truly reach zero in finite time, but it gets vanishingly small.

To find the horizontal position $x$, we integrate the velocity:

$$ x = \int_0^t v_x dt = \int_0^t u_x e^{-kt} dt $$

Evaluating this integral yields the final horizontal position:

$$ \mathbf{x = \frac{u_x}{k}(1 - e^{-kt})} $$

### Conquering the Vertical Motion

Now, let's face the vertical motion. The differential equation here is slightly more complex:

$$ \frac{dv_y}{dt} = -k v_y - g $$

We separate the variables again to prepare for integration:

$$ \frac{dv_y}{k v_y + g} = -dt $$

Integrating from the initial vertical velocity $u_y = u\sin	heta$ to $v_y$, we get:

$$ \frac{1}{k} \ln\left(\frac{k v_y + g}{k u_y + g}ight) = -t $$

Solving for $v_y$, we uncover the vertical velocity function:

$$ \mathbf{v_y = \left(u_y + \frac{g}{k}ight)e^{-kt} - \frac{g}{k}} $$

To find the vertical position $y$, we integrate $v_y$ with respect to time. It requires a bit of algebraic stamina, but the integration is straightforward:

$$ y = \int_0^t \left[ \left(u_y + \frac{g}{k}ight)e^{-kt} - \frac{g}{k} ight] dt $$

After plugging in the limits, we arrive at the final expression for the vertical coordinate:

$$ \mathbf{y = \left(\frac{u_y}{k} + \frac{g}{k^2}ight)(1 - e^{-kt}) - \frac{g}{k}t} $$

### The Endgame: Terminal Velocity

Finally, what happens to the marble after a very long time? As $t 	o \infty$, the exponential term $e^{-kt}$ approaches zero.

The horizontal velocity $v_x$ becomes exactly zero. The marble completely stops moving forward.

But look at the vertical velocity! As the exponential term vanishes, $v_y$ approaches a constant value of $-\frac{g}{k}$.

This is the **terminal velocity**. The negative sign simply indicates that the marble is falling downwards at a constant speed, perfectly balancing the force of gravity with the upward drag force.

Visualized Solution

Initial Setup: $u$

Net Acceleration $a$

Resolving Components

Horizontal Velocity $v_{x}$

Horizontal Position $x$

Vertical Velocity $v_{y}$

Vertical Position $y$

Terminal Velocity

Food for Thought

The Sigma Insight: Motion in a Plane

The Setup

Divide and Conquer

Taming the Horizontal Motion

Conquering the Vertical Motion

The Endgame

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Visualized Solution

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

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