Animated Solution for Physics - Kinematics: Comprehension Passage

Have you ever looked at a graph and realized it's telling you a complete, action-packed story? This problem is a beautiful example of how a simple curve can hide the entire physical reality of two stones falling from a tower. Let's embark on this journey and decode the graph, piece by piece.

Imagine standing at the top of a tall tower. You drop the first stone. As it falls, gravity accelerates it downwards. The distance it covers is given by the classic kinematic equation, $y_1=\frac{1}{2}g{t}^2$. Taking $g=10{\text{ m/s}}^2$, this simplifies to $y_1=5t^2$.

Now, what about the second stone? It's still in your hand! So, the separation $s$ between the two stones is exactly the distance the first stone has fallen. Therefore, $s=5t^2$. Because the separation depends on the square of time, the graph of $s$ versus $t$ is a parabola. This perfectly explains the first curved portion of our graph, from the origin O to point A.

Look closely at the graph. At exactly $t=1\text{ s}$, the shape of the graph abruptly changes from a parabola to a straight line. In physics, a sudden change in a graph's behavior usually means a sudden change in the physical conditions. This is the exact moment you let go of the second stone!

At $t=1\text{ s}$, the separation is $s=5(1{)}^2=5\text{ m}$. So, point A on our graph has the coordinates $(1,5)$.

Now things get really interesting. For $t>1\text{ s}$, both stones are in free fall. The first stone has been falling for a time $t$, so its distance is still $y_1=5t^2$. But the second stone started a second late. So, the time it has been falling is $(t-1)$. Its distance from the top is $y_2=5(t-1{)}^2$.

To find the separation between them, we subtract their distances: Let's expand the squared term:

Look at that! The $t^2$ terms completely cancel out. We are left with a beautiful linear equation. This is why the separation grows at a constant rate, and why the portion AB on the graph is a perfectly straight line.

But this straight line doesn't go on forever. It ends abruptly at point B, where $t=3\text{ s}$. Why would the linear relationship break? Because one of the stones is no longer in free fall! At $t=3\text{ s}$, the first stone smashes into the ground.

This is a massive clue. If the first stone was dropped at $t=0$ and hit the ground at $t=3\text{ s}$, we can easily find the height of the tower. The total distance it fell is: And just like that, we've solved the first part of the problem! The tower is 45 meters tall. We can also find the separation at this instant: $s=10(3)-5=25\text{ m}$. So, point B is at $(3,25)$.

Just to be absolutely rigorous, let's look at the final part of the graph. After $t=3\text{ s}$, the first stone is resting on the ground, 45 meters below the top. The second stone is still falling. The separation is now the distance from the ground to the second stone: This is an inverted parabola, which perfectly matches the portion BC. The second stone will hit the ground when the separation becomes zero. Setting $s=0$, we get $5(t-1{)}^2=45$, which gives $t=4\text{ s}$. This matches point C on our graph perfectly!

The second question asks for the state of the second stone exactly when the first stone hits the ground. We know this happens at $t=3\text{ s}$.

Since the second stone was dropped at $t=1\text{ s}$, it has been falling for exactly $3-1=2\text{ s}$. Its velocity is simply $v=gt$: The distance it has fallen from the top is: Since the tower is 45 meters tall, its height above the ground is:

So, the second stone is moving at $20\text{ m/s}$ at a height of $25\text{ m}$.

By carefully reading the story hidden in the graph, we've completely reconstructed the physical event. This is the true power of graphical analysis in kinematics!