In physics, problems involving objects moving under gravity (projectile motion or freefall) are fundamental to understanding kinematics. This scenario presents a classic example: two objects starting their journeys at the same time from different points and in different directions, all while being influenced by the Earth's gravitational pull. We'll break down how to determine when and where these objects will meet.
The Problem Statement
An object is dropped from the top of a tower of height $156.8 \text{ m}$. At the same time, another object is thrown vertically upward from the foot of the tower with an initial velocity of $78.1 \text{ m/s}$. We need to determine when and where these two objects will meet.
Key Concepts and Principles
To solve this problem, we'll use the equations of motion for constant acceleration, specifically for objects under gravity. The key principles are:
- Constant Acceleration: The only significant acceleration acting on both objects (ignoring air resistance) is due to gravity, $g$, which is approximately $9.8 \text{ m/s}^2$ downwards.
- Kinematic Equations: We'll use the position-time equation:
$s = s_0 + v_0 t + \frac{1}{2} a t^2$
Where:
- $s$ is the final position
- $s_0$ is the initial position
- $v_0$ is the initial velocity
- $t$ is the time elapsed
- $a$ is the constant acceleration
- Coordinate System: It's crucial to define a consistent coordinate system. We'll set the ground as $y=0$ and upward as the positive $y$-direction. Therefore, acceleration due to gravity ($a = -g$) will be negative.
- Meeting Condition: The objects meet when their positions ($y$) are the same at the same time ($t$).
Step-by-Step Solution
Let's break down the problem for each object.
1. Identify Knowns
- Height of the tower, $H = 156.8 \text{ m}$
- Initial velocity of the dropped object (Object 1), $u_1 = 0 \text{ m/s}$
- Initial velocity of the object thrown upward (Object 2), $u_2 = 78.1 \text{ m/s}$
- Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$
2. Set Up Equations of Motion for Each Object
We'll use the coordinate system where the ground is $y=0$ and upward is positive.
For Object 1 (Dropped from Top)
- Initial position, $y_{1,0} = H = 156.8 \text{ m}$
- Initial velocity, $u_1 = 0 \text{ m/s}$
- Acceleration, $a_1 = -g = -9.8 \text{ m/s}^2$
The position of Object 1 at time $t$ is: $y_1(t) = y_{1,0} + u_1 t + \frac{1}{2} a_1 t^2$ $y_1(t) = 156.8 + (0)t + \frac{1}{2} (-9.8) t^2$ $y_1(t) = 156.8 - 4.9 t^2 \quad \ldots (1)$
For Object 2 (Thrown Upward from Foot)
- Initial position, $y_{2,0} = 0 \text{ m}$
- Initial velocity, $u_2 = 78.1 \text{ m/s}$
- Acceleration, $a_2 = -g = -9.8 \text{ m/s}^2$
The position of Object 2 at time $t$ is: $y_2(t) = y_{2,0} + u_2 t + \frac{1}{2} a_2 t^2$ $y_2(t) = 0 + (78.1)t + \frac{1}{2} (-9.8) t^2$ $y_2(t) = 78.1 t - 4.9 t^2 \quad \ldots (2)$
3. Determine When They Meet (Solve for Time, $t$)
The objects meet when their positions are equal: $y_1(t) = y_2(t)$.
$156.8 - 4.9 t^2 = 78.1 t - 4.9 t^2$
Notice that the gravitational acceleration terms ($-4.9 t^2$) cancel out on both sides. This is a common and intuitive simplification in problems where both objects are under the same constant acceleration. Their relative acceleration is zero.
$156.8 = 78.1 t$
Now, solve for $t$: $t = \frac{156.8}{78.1}$ $t \approx 2.00768 \text{ s}$
Rounding to three significant figures (due to $78.1 \text{ m/s}$): $t \approx 2.01 \text{ s}$
So, the objects meet approximately $2.01$ seconds after they begin their motion.
4. Determine Where They Meet (Solve for Position, $y$)
To find the meeting point, substitute the value of $t$ back into either equation (1) or (2). Let's use equation (2) for Object 2, as it represents the height from the ground directly.
$y_{meet} = 78.1 t - 4.9 t^2$ $y_{meet} = 78.1 (2.00768) - 4.9 (2.00768)^2$
First term: $78.1 \times 2.00768 \approx 156.8 \text{ m}$ (This is expected since $H = u_2 t$ was derived earlier)
Second term: $4.9 \times (2.00768)^2 = 4.9 \times 4.03078 \approx 19.75 \text{ m}$
Now, calculate $y_{meet}$: $y_{meet} = 156.8 - 19.75$ $y_{meet} = 137.05 \text{ m}$
Rounding to three significant figures: $y_{meet} \approx 137 \text{ m}$
The objects meet at a height of approximately $137 \text{ m}$ from the ground.
Intuition and Discussion
- Relative Motion: The cancellation of the $4.9 t^2$ terms is a powerful insight. It means that from the perspective of an observer moving with the average acceleration of both objects (which is just $g$ downwards), the problem simplifies to one of constant relative velocity. The relative velocity of the dropped object with respect to the upward-thrown object is $u_1 - u_2 = 0 - 78.1 = -78.1 \text{ m/s}$ (if positive is up). Or, more simply, the objects are closing the distance between them at a rate equal to the sum of their initial speeds if we consider relative speed in a non-accelerating frame, but under gravity, it simplifies even more. The crucial point is that gravity affects both objects equally, so it doesn't change their relative velocity or position in a constant acceleration problem.
- Total Distance Covered: The sum of the distances covered by the two objects must equal the total height of the tower. Let $d_1$ be the distance fallen by the first object and $d_2$ be the distance risen by the second object from its starting point. Then $d_1 + d_2 = H$. Our solution finds the meeting height from the ground ($y_{meet}$). The distance fallen by the first object is $H - y_{meet} = 156.8 - 137.05 = 19.75 \text{ m}$. The distance risen by the second object is $y_{meet} = 137.05 \text{ m}$. Their sum is $19.75 + 137.05 = 156.8 \text{ m}$, which matches the tower's height. This consistency check confirms our result.
Final Answer
The two objects will meet approximately $2.01$ seconds after their release, at a height of approximately $137$ meters from the foot of the tower.