Drag force of a fluid on a body is proportional to the velocity of the body relative to the fluid. A

Q: Why does the velocity approach a constant value instead of increasing forever?

As the stone speeds up, the upward drag force increases. Eventually, this drag force becomes exactly equal to the downward force of gravity. At this point, the net force is zero, so the stone stops accelerating and moves at a constant 'terminal' velocity.

Q: Can a stone really slow down when falling through water?

Yes! If you drop a stone from a very high bridge, it might be moving extremely fast when it hits the water—faster than its terminal velocity in water. Because the drag force at that high speed is greater than gravity, the net force is upwards, causing the stone to decelerate until it reaches terminal velocity.

Q: What does the horizontal line in Graph 3 mean physically?

It means the stone entered the water at exactly its terminal velocity. Since the forces were perfectly balanced the moment it touched the water, it didn't need to speed up or slow down; it just continued at that steady pace.

Q: Why is the drag force proportional to velocity?

For small objects moving relatively slowly through a viscous fluid (like a small stone in water), the flow is laminar, and Stokes' Law applies, making drag proportional to velocity ($F_d = -kv$). For larger objects or higher speeds, drag is often proportional to velocity squared ($v^2$). The problem specifically states it's proportional to $v$.

Q: How do we know Graph 1 is for a large height and not a small height?

A large drop height means the stone has more time to accelerate in the air, so it hits the water with a high speed. Graph 1 starts at a high speed (above $v_0$) and slows down, which perfectly matches a stone entering the water faster than its terminal velocity.

Question

Drag force of a fluid on a body is proportional to the velocity of the body relative to the fluid. A student drops several small identical stones from different heights over a deep lake and prepares graphs between speed $v$ of every stone in the water and time $t$. The graphs can be divided into three categories as shown here. Which of the following explanations of these graphs appear reasonable?

Options:

(A) The first graph is for a stone dropped from a small height, and the second is for a stone dropped from a large height.

(B) The first graph is for a stone dropped from a large height, and the second is for a stone dropped from a small height.

(C) The third graph is for a stone dropped from a height sufficient to acquire the speed $v_0$ at the instant it enters the water.

(D) The third graph is not possible.

Sigma Percentile · Accepted Answer

Imagine standing on a bridge over a deep, calm lake. You hold a small stone in your hand and let it drop. As it falls through the air, gravity accelerates it, and it gains speed. But the moment it pierces the surface of the water, the rules of the game change dramatically. The water pushes back. This pushback is the **drag force**, and in this problem, we are told it is directly proportional to the stone's velocity.

### The Math of Drag

Let's translate this physical intuition into mathematics. Once the stone is submerged, there are two primary forces acting on it: the downward pull of gravity ($mg$) and the upward drag force ($F_d = kv$). According to Newton's Second Law, the net force dictates the acceleration:

$$ma = mg - kv$$

Dividing by the mass $m$, we get an expression for the acceleration:

$$a = g - \frac{k}{m}v$$

This equation is the master key to the whole problem. Notice how the acceleration depends on the velocity $v$. As the stone speeds up, the term $\frac{k}{m}v$ grows larger, which means the overall acceleration $a$ gets smaller.

### The Concept of Terminal Velocity

Eventually, if the stone falls long enough, the upward drag force will perfectly balance the downward gravitational force. When this happens, the net force becomes zero, and the acceleration drops to zero. The constant speed the stone achieves at this point is called the **terminal velocity**, denoted as $v_0$.

Setting $a = 0$ in our equation gives:

$$0 = g - \frac{k}{m}v_0 \implies v_0 = \frac{mg}{k}$$

We can use this elegant result to rewrite our acceleration equation. By substituting $\frac{k}{m} = \frac{g}{v_0}$, we get:

$$a = \frac{g}{v_0}(v_0 - v)$$

This form is incredibly powerful. It tells us immediately that the sign of the acceleration depends entirely on whether the current velocity $v$ is greater than or less than the terminal velocity $v_0$.

### Decoding the Graphs

Now, let's look at the three graphs provided and connect them to our physical scenarios.

**Case 1: The High Drop (Graph 1)**
Suppose you drop the stone from a very large height. By the time it hits the water, it has accelerated so much in the air that its entry speed $u$ is *greater* than the terminal velocity in water ($u > v_0$).

Looking at our master equation, if $v > v_0$, the term $(v_0 - v)$ is negative, making the acceleration $a$ negative. The drag force is actually stronger than gravity! The stone will rapidly decelerate, and its velocity curve will decay downwards, asymptotically approaching $v_0$. This perfectly matches **Graph 1**.

**Case 2: The Low Drop (Graph 2)**
What if you drop the stone from just a few inches above the water? Its entry speed $u$ will be very low, certainly less than the terminal velocity ($u < v_0$).

Here, $(v_0 - v)$ is positive, so the acceleration $a$ is positive. Gravity is winning the tug-of-war. The stone will continue to speed up in the water, but at a decreasing rate, until its velocity levels off at $v_0$. This upward curve is exactly what we see in **Graph 2**.

**Case 3: The Perfect Drop (Graph 3)**
Finally, imagine dropping the stone from a very specific, calculated height. It falls through the air and hits the water with an entry speed exactly equal to the terminal velocity ($u = v_0$).

In this magical scenario, $(v_0 - v) = 0$, so the acceleration is zero from the very first millisecond. The forces are perfectly balanced upon entry, and the stone simply cruises downward at a constant speed $v_0$. This horizontal line is depicted in **Graph 3**.

### Conclusion

By analyzing the physics, we've deduced that Graph 1 corresponds to a large drop height, Graph 2 to a small drop height, and Graph 3 to a specific height that grants exactly the terminal velocity upon entry. Therefore, the explanations given in options (b) and (c) are both perfectly reasonable and correct.

Select Answer:

Visualized Solution

Forces on the Stone

Equation of Motion

Terminal Velocity ( $v_{0}$ )

Acceleration in terms of $v_{0}$

Analyzing Graph 1

Analyzing Graph 2

Analyzing Graph 3

Final Conclusion

The Sigma Insight: Newton's Laws of Motion

The Math of Drag

The Concept of Terminal Velocity

Decoding the Graphs

Conclusion

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