In physics, problems involving objects moving under the influence of gravity are a cornerstone of kinematics. This particular challenge presents a scenario where two objects begin their motion simultaneously from different points on a tower, moving towards each other. Understanding how to apply the equations of motion and recognizing the power of relative motion simplifies these kinds of problems significantly.
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 the two objects meet.
Understanding the Scenario
We have two objects in vertical motion:
- Object 1 (Dropped): Starts from the top of the tower, initial velocity is $0 \text{ m/s}$, and accelerates downwards due to gravity.
- Object 2 (Thrown Up): Starts from the base of the tower, has an initial upward velocity of $78.1 \text{ m/s}$, and also accelerates downwards due to gravity.
They start at the same time and are moving towards each other. Our goal is to find the time ($t$) at which their vertical positions are identical and that specific height ($y$) above the ground.
Key Concepts and Intuition: Relative Motion
Before diving into the full equations, let's build some intuition using the concept of relative motion. Both objects are subject to the same acceleration due to gravity, $g$, acting downwards. This is a crucial observation!
If we consider the motion of Object 1 relative to Object 2 (or vice-versa), the acceleration term cancels out. Why?
Let's denote the position of Object 1 as $y_1(t)$ and Object 2 as $y_2(t)$.
$y_1(t) = y_{1,0} + v_{1,0}t + \frac{1}{2}at^2$ $y_2(t) = y_{2,0} + v_{2,0}t + \frac{1}{2}at^2$
If the acceleration $a$ is the same for both objects (in this case, $a = -g$), then their relative position $\Delta y(t) = y_1(t) - y_2(t)$ becomes:
$\Delta y(t) = (y_{1,0} - y_{2,0}) + (v_{1,0} - v_{2,0})t + \frac{1}{2}(a-a)t^2$ $\Delta y(t) = \Delta y_0 + v_{rel,0}t$
This shows that the relative motion between the two objects is one of constant velocity! The acceleration due to gravity does not affect their relative positions.
In our problem:
- Initial separation $\Delta y_0 = H - 0 = H = 156.8 \text{ m}$.
- Initial velocity of object 1 (downwards) $v_{1,0} = 0 \text{ m/s}$.
- Initial velocity of object 2 (upwards) $v_{2,0} = 78.1 \text{ m/s}$.
The closing speed is the sum of their absolute speeds if they were moving towards each other. However, with our chosen coordinate system (upwards positive), the relative velocity needs careful consideration.
Let's stick to the positions for now. When they meet, $y_1(t) = y_2(t)$.
$y_{1,0} + v_{1,0}t - \frac{1}{2}gt^2 = y_{2,0} + v_{2,0}t - \frac{1}{2}gt^2$
As you can see, the $gt^2$ terms cancel out, leaving:
$y_{1,0} + v_{1,0}t = y_{2,0} + v_{2,0}t$
Rearranging to find time:
$y_{1,0} - y_{2,0} = (v_{2,0} - v_{1,0})t$ $H = (v_{2,0} - v_{1,0})t$
This is the initial separation divided by the initial relative velocity (the rate at which they are closing the gap). Since $v_{1,0}=0$, this simplifies to:
$H = v_{2,0}t$
This powerful simplification allows us to find the time of meeting very quickly.
Setting Up the Equations (Formal Approach)
Let's define our coordinate system:
- Origin ($y=0$) at the foot of the tower.
- Upward direction is positive ($+y$).
- Acceleration due to gravity $g = 9.8 \text{ m/s}^2$ (acting downwards, so $a = -9.8 \text{ m/s}^2$).
Given values:
- Height of the tower, $H = 156.8 \text{ m}$
- Initial velocity of object 2, $v_{2,0} = 78.1 \text{ m/s}$
For Object 1 (Dropped from Top)
- Initial position: $y_{1,0} = H = 156.8 \text{ m}$
- Initial velocity: $v_{1,0} = 0 \text{ m/s}$ (since it's dropped)
- Acceleration: $a_1 = -g = -9.8 \text{ m/s}^2$
The equation for its position $y_1(t)$ at time $t$ is:
$y_1(t) = y_{1,0} + v_{1,0}t + \frac{1}{2}a_1t^2$ $y_1(t) = 156.8 + (0)t + \frac{1}{2}(-9.8)t^2$ $y_1(t) = 156.8 - 4.9t^2 \quad \text{(Equation 1)}$
For Object 2 (Thrown Up from Foot)
- Initial position: $y_{2,0} = 0 \text{ m}$
- Initial velocity: $v_{2,0} = 78.1 \text{ m/s}$ (upward, so positive)
- Acceleration: $a_2 = -g = -9.8 \text{ m/s}^2$
The equation for its position $y_2(t)$ at time $t$ is:
$y_2(t) = y_{2,0} + v_{2,0}t + \frac{1}{2}a_2t^2$ $y_2(t) = 0 + (78.1)t + \frac{1}{2}(-9.8)t^2$ $y_2(t) = 78.1t - 4.9t^2 \quad \text{(Equation 2)}$
Solving for Time (When They Meet)
The objects meet when their positions are the same, i.e., $y_1(t) = y_2(t)$.
Set Equation 1 equal to Equation 2:
$156.8 - 4.9t^2 = 78.1t - 4.9t^2$
Notice that the $-4.9t^2$ term cancels out on both sides, as predicted by our relative motion intuition!
$156.8 = 78.1t$
Now, solve for $t$:
$t = \frac{156.8}{78.1}$ $t \approx 2.007682458 \text{ s}$
Rounding to two decimal places, which is appropriate given the precision of the input values:
$\boxed{t \approx 2.01 \text{ s}}$
Solving for Position (Where They Meet)
To find where they meet, substitute the value of $t$ into either Equation 1 or Equation 2. Let's use Equation 2 for Object 2's position, as it directly gives the height above the ground.
$y_2(t) = 78.1t - 4.9t^2$
Using the more precise fractional value for $t$ to maintain accuracy:
$y_{meet} = 78.1\left(\frac{156.8}{78.1}\right) - 4.9\left(\frac{156.8}{78.1}\right)^2$ $y_{meet} = 156.8 - 4.9\left(\frac{156.8}{78.1}\right)^2$ $y_{meet} = 156.8 - 4.9(2.007682458)^2$ $y_{meet} = 156.8 - 4.9(4.030799738)$ $y_{meet} = 156.8 - 19.7509187$ $y_{meet} \approx 137.0490813 \text{ m}$
Rounding to two decimal places:
$\boxed{y_{meet} \approx 137.05 \text{ m}}$
This height is measured from the foot of the tower.
Summary of Results
The two objects will meet:
- When: Approximately $2.01 \text{ seconds}$ after they start their motion.
- Where: Approximately $137.05 \text{ meters}$ above the foot of the tower.
Conclusion
This problem beautifully illustrates the principles of kinematics under constant acceleration. By setting up the equations of motion for each object and recognizing that they meet when their positions are equal, we can solve for both the time and location of their rendezvous. The intuitive understanding gained from relative motion, where the acceleration due to gravity cancels out, provides an elegant shortcut to finding the time of collision, emphasizing that physics problems often have multiple paths to a solution.