A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes the pot 0.400 s to pass from the top to the bottom of this window, which is 2.00 m high. How far is the top of the window below the windowsill from which the flowerpot fell?

We are tasked with determining the distance between the windowsill and the top of the window below, given the time it takes for a flowerpot to fall past a window and the height of that window.

### Step 1: Identify known values.
- Time taken to pass the window (\( t \)): 0.400 s
- Height of the window (\( h \)): 2.00 m
- Acceleration due to gravity (\( g \)): \( 9.8 \, \text{m/s}^2 \)
- The initial velocity (\( v_1 \)) of the pot when it reaches the top of the window is unknown.
- The final velocity (\( v_2 \)) of the pot when it reaches the bottom of the window is also unknown.

### Step 2: Use kinematics to find the velocity at the top and bottom of the window.

Using the equation for constant acceleration:

\[
h = \frac{1}{2} (v_1 + v_2)t
\]

We can rearrange this equation to find the average velocity while passing the window:

\[
v_{\text{avg}} = \frac{h}{t} = \frac{2.00 \, \text{m}}{0.400 \, \text{s}} = 5.00 \, \text{m/s}
\]

Since the average velocity is the midpoint between the initial and final velocities:

\[
v_{\text{avg}} = \frac{v_1 + v_2}{2}
\]

Therefore,

\[
v_1 + v_2 = 2 \times v_{\text{avg}} = 2 \times 5.00 \, \text{m/s} = 10.00 \, \text{m/s}
\]

### Step 3: Use kinematic equations to relate velocity and acceleration.

From the kinematic equation:

\[
v_2 = v_1 + g t
\]

Substitute \( t = 0.400 \, \text{s} \) and \( g = 9.8 \, \text{m/s}^2 \):

\[
v_2 = v_1 + 9.8 \times 0.400
\]
\[
v_2 = v_1 + 3.92
\]

Now substitute \( v_2 = 10.00 - v_1 \) (from \( v_1 + v_2 = 10.00 \)):

\[
10.00 - v_1 = v_1 + 3.92
\]
\[
10.00 = 2v_1 + 3.92
\]
\[
2v_1 = 10.00 - 3.92
\]
\[
2v_1 = 6.08
\]
\[
v_1 = 3.04 \, \text{m/s}
\]

Thus, the initial velocity at the top of the window is \( v_1 = 3.04 \, \text{m/s} \).

### Step 4: Find the distance from the windowsill to the top of the window.

Now, we will use the kinematic equation to find the distance \( d \) from the windowsill (where the velocity is \( v_0 = 0 \)) to the top of the window (where the velocity is \( v_1 = 3.04 \, \text{m/s} \)):

\[
v_1^2 = v_0^2 + 2 g d
\]
\[
(3.04)^2 = 0 + 2 \times 9.8 \times d
\]
\[
9.24 = 19.6 \times d
\]
\[
d = \frac{9.24}{19.6} = 0.471 \, \text{m}
\]

### Final Answer:
The top of the window is approximately **0.471 m** below the windowsill.

Would you like more details on this solution, or do you have any questions?

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**5 related questions:**
1. How would the result change if air resistance was considered?
2. What if the height of the window was different, say 3 meters?
3. How would the problem change if the object was thrown downwards?
4. Can you determine the total time for the pot to hit the ground?
5. What is the velocity of the pot when it hits the ground?

**Tip:** Always double-check your units to ensure consistency when solving physics problems.

Solve a kinematics problem where a flowerpot falls off a windowsill, passing a window in 0.400 seconds. Using kinematic equations, determine the distance from the windowsill to the top of the window. This problem covers concepts such as free fall and constant acceleration, suitable for students in Grades 10-12.

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