Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10

Q: Why does the graph start as a straight line instead of a curve?

Because both stones are in free fall, they have the same acceleration (-g). Their relative acceleration is zero. When relative acceleration is zero, relative velocity is constant, which means the relative position changes linearly with time.

Q: Why does the shape of the graph change at t = 8 seconds?

At t = 8 seconds, the first stone hits the ground and stops moving. It is no longer in free fall, so its acceleration changes from -g to 0. This abrupt change in the state of the first stone causes the relative motion equation to change.

Q: How do we know the curve after 8 seconds is a downward parabola and not an upward one?

After 8 seconds, the relative position is just the position of the second stone: y = 240 + 40t - 5t^2. The coefficient of the t^2 term is negative (-5). Mathematically, a quadratic equation with a negative leading coefficient always forms a parabola that opens downwards.

Q: What happens to the graph after t = 12 seconds?

At t = 12 seconds, the second stone also hits the ground. After this point, both stones are at rest on the ground (y1 = 0, y2 = 0). Therefore, their relative position becomes zero and stays zero. The graph would just be a flat line on the x-axis.

Q: Would the graph be different if the stones were thrown downwards?

Yes! If thrown downwards, they would hit the ground much sooner. The initial straight-line phase would be shorter, but the fundamental logic remains the same: a straight line while both are falling, followed by a parabola when one hits the ground before the other.

Question

Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? Assume stones do not rebound after hitting the ground and neglect air resistance, take $g = 10 	ext{ m/s}^2$ )

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Stones

Imagine standing on the edge of a $240 	ext{ m}$ high cliff. You hold two stones, one in each hand. You throw both of them straight up at the exact same moment. Stone 1 is tossed gently with an initial velocity of $10 	ext{ m/s}$, while Stone 2 is hurled with a much greater force, leaving your hand at $40 	ext{ m/s}$.

Our goal is to plot the relative position of Stone 2 with respect to Stone 1, which is mathematically expressed as $y_2 - y_1$, against time $t$. To do this, we must first understand the individual journey of each stone.

### Phase 1: The Dance in the Air

Let's set the ground as our origin ($y = 0$). The initial height for both stones is $y_0 = 240 	ext{ m}$. Using the second equation of motion, $y = y_0 + ut - \frac{1}{2}gt^2$, we can write the position equations for both stones:

$$ y_1(t) = 240 + 10t - 5t^2 $$

$$ y_2(t) = 240 + 40t - 5t^2 $$

As long as both stones are freely flying in the air, they experience the exact same gravitational acceleration ($g = 10 	ext{ m/s}^2$ downwards). Because their accelerations are identical, their relative acceleration is zero!

Let's see what happens when we calculate their relative position:

$$ y_{rel} = y_2(t) - y_1(t) $$

$$ y_{rel} = (240 + 40t - 5t^2) - (240 + 10t - 5t^2) $$

$$ y_{rel} = 30t $$

The initial height and the gravity terms perfectly cancel out. The relative position grows linearly at a rate of $30 	ext{ m/s}$ (which is their relative velocity). Therefore, the graph starts as a straight line passing through the origin.

### The Turning Point: Stone 1 Bows Out

This linear relationship is beautiful, but it doesn't last forever. Stone 1 was thrown with a smaller velocity, so it will inevitably hit the ground first. We need to find the exact moment this happens. We set the position of Stone 1 to zero:

$$ 240 + 10t - 5t^2 = 0 $$

Dividing the entire equation by $-5$, we get a neat quadratic equation:

$$ t^2 - 2t - 48 = 0 $$

Factoring this yields $(t-8)(t+6) = 0$. Since time cannot be negative, we find that Stone 1 hits the ground at exactly $t = 8 	ext{ s}$. At this instant, the relative position is $30 	imes 8 = 240 	ext{ m}$.

### Phase 2: The Solo Flight

After $8 	ext{ seconds}$, Stone 1 is lying motionless on the ground. Its position is permanently $y_1 = 0$. However, Stone 2 was thrown much faster and is still completing its trajectory.

For $t > 8 	ext{ s}$, the relative position equation changes abruptly. It is now simply the position of Stone 2:

$$ y_{rel} = y_2(t) - 0 $$

$$ y_{rel} = 240 + 40t - 5t^2 $$

Look closely at this new equation. It is a quadratic function of time, and the coefficient of the $t^2$ term is negative ($-5$). In coordinate geometry, a quadratic equation with a negative leading coefficient represents a parabola that opens downwards.

Therefore, after $t = 8 	ext{ s}$, the graph transitions from a straight line to a downward-facing curve. It will continue along this parabolic path until Stone 2 also hits the ground (which happens at $t = 12 	ext{ s}$).

Matching this two-part behavior—a straight line followed by a downward parabola—with our given options, we can confidently conclude that **Option (b)** is the correct representation of the motion.

Select Answer:

Visualized Solution

$Initial Setup$

$Equations of Motion$

$Phase 1: Both in Air$

$Time of Flight for Stone 1$

$Phase 2: Stone 1 at Rest$

$Analyzing the Curve$

$Final Conclusion$

The Sigma Insight: Relative Velocity

The Setup

Phase 1

The Turning Point

Phase 2

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List-I

List-II