A dog sees a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50 s, and the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go ?

Answer

To solve the problem of how high above the window top the flowerpot goes, we can follow these steps:

### Step 1: Understand the Motion
The flowerpot moves upward and then downward, passing by the window. The total time it is in view is 0.50 seconds, and the height of the window is 2.00 meters.

### Step 2: Determine the Initial Conditions
When the flowerpot passes the top of the window, we can assume its initial velocity (u) is 0 m/s at the highest point of its trajectory. The flowerpot then falls down past the window.

### Step 3: Calculate the Final Velocity at the Bottom of the Window
Using the kinematic equation:
\[ v^2 = u^2 + 2gh \]
Where:
- \( v \) = final velocity
- \( u \) = initial velocity (0 m/s at the top)
- \( g \) = acceleration due to gravity (approximately 9.8 m/s²)
- \( h \) = height of the window (2 m)

Substituting the values:
\[ v^2 = 0 + 2 \times 9.8 \times 2 \]
\[ v^2 = 39.2 \]
\[ v = \sqrt{39.2} \approx 6.26 \, \text{m/s} \]

### Step 4: Calculate the Time to Fall Through the Window
The time to fall through the window can be calculated using the formula:
\[ t = \frac{h}{v} \]
Where \( h \) is the height of the window (2 m) and \( v \) is the velocity at the bottom of the window (6.26 m/s).
\[ t = \frac{2}{6.26} \approx 0.319 \, \text{s} \]

### Step 5: Determine the Time to Reach the Highest Point
Since the total time the flowerpot is in view is 0.50 seconds, we can find the time taken to reach the highest point:
\[ t_{\text{up}} = 0.50 - t_{\text{fall}} \]
\[ t_{\text{up}} = 0.50 - 0.319 \approx 0.181 \, \text{s} \]

### Step 6: Calculate the Maximum Height Above the Window
Using the time to reach the highest point, we can calculate the height above the window using the formula:
\[ h_{\text{max}} = u t + \frac{1}{2} g t^2 \]
Where:
- \( u = 6.26 \, \text{m/s} \) (initial velocity at the bottom of the window)
- \( t = 0.181 \, \text{s} \)

Substituting the values:
\[ h_{\text{max}} = 6.26 \times 0.181 + \frac{1}{2} \times 9.8 \times (0.181)^2 \]
\[ h_{\text{max}} = 1.134 + 0.161 \approx 1.295 \, \text{m} \]

### Final Step: Total Height Above the Window
Since the height of the window is 2 m, the total height above the window is:
\[ h_{\text{total}} = h_{\text{max}} + h_{\text{window}} \]
\[ h_{\text{total}} = 1.295 + 2 \approx 3.295 \, \text{m} \]

### Conclusion
The flowerpot goes approximately 1.295 meters above the top of the window.

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