To study effect of air resistance, a rubber ball was shot vertically upwards from a spring gun from

Q: Why did we assume air resistance is zero when the question explicitly says it's an experiment to study it?

The question describes the *intent* of the experiment. However, the *result* of the experiment is the graph itself. Since the graph is a perfect straight line, the upward acceleration equals the downward acceleration. Physically, this is only possible if air resistance is negligible ($a_{up} = g + a_r$, $a_{down} = g - a_r$). The graph tells us the truth!

Q: How do we know the graph ends exactly when the ball hits the ground?

The problem asks for the speed when it strikes the ground, implying the recorded data covers the entire flight. The graph abruptly ends at a specific point ($t=4.5$ s), which logically corresponds to the end of the motion, i.e., hitting the ground.

Q: Why did we take $g = 10 \text{ m/s}^2$ instead of $9.8 \text{ m/s}^2$?

In JEE problems, unless specified otherwise, $g$ is often taken as $10 \text{ m/s}^2$ to keep calculations clean. If you use $9.8$, the scale becomes $4.9 \text{ m/s}$ per division, leading to a final speed of $24.5 \text{ m/s}$, which is usually rounded to 25 in such contexts, but 10 makes the grid perfectly integer-aligned.

Q: What does the '20th floor' information do in this problem?

It's actually a distractor! You don't need to know the height of the building to solve the problem. The kinematics are fully encoded in the velocity-time graph. Sometimes, extra information is given to test if you can identify what's truly necessary.

Q: How did you count the divisions on the x-axis so confidently?

By looking at the labels '1' and '2'. There are exactly two grid squares between 0 and 1, and two between 1 and 2. This establishes that one grid square on the x-axis represents 0.5 seconds. Counting the squares after $t=2$ leads us to the end point at $t=4.5$ s.

Question

To study effect of air resistance, a rubber ball was shot vertically upwards from a spring gun from $20^{	ext{th}}$ floor of a tall building. Velocity of the ball was recorded at regular intervals of time and the data obtained were plotted on a graph paper. Some of the marking on the axes are erased as shown in the following figure. With what speed did the ball strike the ground?

Sigma Percentile · Accepted Answer

Imagine you are standing on the 20th floor of a tall building, holding a spring gun. You shoot a rubber ball straight up into the air to study the effects of air resistance. You meticulously record its velocity and plot it on a graph. But oh no! Some of the markings on the y-axis get erased. How do you find the speed at which it hits the ground?

This problem is a beautiful test of your graph-reading skills and physical intuition. Let's dive into it.

### The Setup: A Deceptive Premise

The problem explicitly mentions that the experiment was conducted "to study the effect of air resistance." Naturally, your brain might immediately jump to complex differential equations or non-constant acceleration formulas. But before we panic, let's look at the data we *do* have: the velocity-time ($v-t$) graph.

### The Master Clue: The Straight Line

Look closely at the graph. It is a single, unbroken, perfect straight line. Why is this so crucial?

In kinematics, the slope of a $v-t$ graph represents acceleration. A straight line means the slope is constant, which in turn means the acceleration is constant throughout the entire flight.

Let's think about the physics. If air resistance ($F_r$) were significant, the acceleration on the way up would be $a_{	ext{up}} = g + \frac{F_r}{m}$ (since both gravity and air resistance point downwards). On the way down, the acceleration would be $a_{	ext{down}} = g - \frac{F_r}{m}$ (since air resistance now points upwards, opposing gravity).

If $a_{	ext{up}}$ and $a_{	ext{down}}$ were different, the graph would have a kink at the highest point ($v=0$). It would be steeper on the left and gentler on the right. But our graph has no kink! The slope is identical on both sides. This leads to a stunning revelation: **the air resistance in this specific experiment turned out to be negligible!** The acceleration is simply $g$ ($10 	ext{ m/s}^2$) the whole way through.

### Unlocking the Y-Axis Scale

Now that we know the physical acceleration is $10 	ext{ m/s}^2$, let's calculate the acceleration from the graph's geometry.

We don't know the scale of the y-axis, so let's measure it in "grid divisions". The ball starts at $4$ divisions above the axis and reaches $v=0$ at $t=2 	ext{ s}$. 
The magnitude of the slope is:
$$a = \frac{\Delta v}{\Delta t} = \frac{4 	ext{ div}}{2 	ext{ s}} = 2 	ext{ div/s}^2$$

We have two expressions for the same physical quantity. Let's equate them:
$$2 	ext{ div/s}^2 = 10 	ext{ m/s}^2$$
$$1 	ext{ div} = 5 	ext{ m/s}$$

Boom! We have just decoded the erased markings on the y-axis. Every single grid line represents $5 	ext{ m/s}$.

### The Final Plunge

To find the striking speed, we need to locate the end of the journey on the graph. By carefully counting the grid squares on the x-axis (where 1 square = $0.5 	ext{ s}$), we can trace the line to its termination point.

The graph ends exactly $5$ divisions below the time axis. 
So, the final velocity is:
$$v_f = -5 	ext{ div}$$

Substitute our newly found scale:
$$v_f = -5 	imes 5 	ext{ m/s} = -25 	ext{ m/s}$$

The negative sign simply indicates the downward direction. The speed, which is the magnitude of velocity, is **$25 	ext{ m/s}$**.

Notice how we didn't even need to use the "20th floor" information? In physics problems, sometimes the most elegant solution comes from trusting the geometry of the data rather than getting lost in the distractors. Always let the graph speak to you!

Enter Numerical Value:

Visualized Solution

$v - t Graph Analysis$

$Calculating Slope$

$Effect of Air Resistance$

$Finding the Scale$

$Final Velocity$

$Striking Speed$

$What if Air Resistance was Present?$

The Sigma Insight: Motion Graphs

The Setup

The Master Clue

Unlocking the Y-Axis Scale

The Final Plunge

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