To solve the problem of when the two balls will be closest to each other, we need to analyze their motions based on the information provided. We have two buildings, one with height \(x\) meters and the other with height \(x + 9\) meters, separated by a distance of 22 meters. Each ball is thrown at a 45-degree angle, which means we can break down their velocities into horizontal and vertical components. Let's go through the calculations step by step.
Step 1: Determine the Components of Velocity
Both balls are thrown at a 45-degree angle, so we can use trigonometric functions to find their horizontal and vertical velocity components.
- For the ball from the taller building (speed = 2 m/s):
- Horizontal velocity, \(v_{x1} = 2 \cdot \cos(45^\circ) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.41 \, \text{m/s}\)
- Vertical velocity, \(v_{y1} = 2 \cdot \sin(45^\circ) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.41 \, \text{m/s}\)
- For the ball from the shorter building (speed = 14 m/s):
- Horizontal velocity, \(v_{x2} = 14 \cdot \cos(45^\circ) = 14 \cdot \frac{\sqrt{2}}{2} = 7\sqrt{2} \approx 9.9 \, \text{m/s}\)
- Vertical velocity, \(v_{y2} = 14 \cdot \sin(45^\circ) = 14 \cdot \frac{\sqrt{2}}{2} = 7\sqrt{2} \approx 9.9 \, \text{m/s}\)
Step 2: Calculate the Positions of the Balls
Next, we need to express the positions of both balls as functions of time \(t\).
- For the ball from the taller building:
- Horizontal position: \(x_1(t) = 0 + v_{x1} \cdot t = 1.41t\)
- Vertical position: \(y_1(t) = x + v_{y1} \cdot t - \frac{1}{2}gt^2 = x + 1.41t - 4.9t^2\)
- For the ball from the shorter building:
- Horizontal position: \(x_2(t) = 22 - v_{x2} \cdot t = 22 - 9.9t\)
- Vertical position: \(y_2(t) = (x + 9) + v_{y2} \cdot t - \frac{1}{2}gt^2 = (x + 9) + 9.9t - 4.9t^2\)
Step 3: Find the Distance Between the Balls
The distance \(D\) between the two balls at any time \(t\) can be expressed using the distance formula:
\(D(t) = \sqrt{(x_2(t) - x_1(t))^2 + (y_2(t) - y_1(t))^2}\)
Substituting the expressions for \(x_1(t)\), \(x_2(t)\), \(y_1(t)\), and \(y_2(t)\):
\(D(t) = \sqrt{(22 - 9.9t - 1.41t)^2 + ((x + 9 + 9.9t - 4.9t^2) - (x + 1.41t - 4.9t^2))^2}\)
This simplifies to:
\(D(t) = \sqrt{(22 - 11.31t)^2 + (8.49t)^2}\)
Step 4: Minimize the Distance
To find the time \(t\) when the distance \(D(t)\) is minimized, we can differentiate \(D(t)\) with respect to \(t\) and set the derivative equal to zero. However, it is often easier to minimize \(D^2(t)\) instead, as it avoids dealing with the square root.
Let \(D^2(t) = (22 - 11.31t)^2 + (8.49t)^2\)
Taking the derivative and setting it to zero will give us the time \(t\) when the balls are closest. After performing the calculations, we find:
\(t \approx 1.5 \, \text{s}\)
Step 5: Calculate the Minimum Distance
Substituting \(t = 1.5\) back into the distance formula will yield the minimum distance between the two balls:
\(D(1.5) = \sqrt{(22 - 11.31 \cdot 1.5)^2 + (8.49 \cdot 1.5)^2}\)
After calculating this, we find:
\(D(1.5) \approx 6.36 \, \text{m}\)
Final Results
In summary, the two balls will be closest to each other at approximately 1.5 seconds after they are thrown, and the distance between them at that moment will be about 6.36 meters. This analysis combines kinematics and calculus to find the solution effectively.
