A flowerpot falls off a windowsill and falls past the window below. It takes the pot 0.42 s to pass this window, which is 1.90 m high. How far is the top of the window below the windowsill from which the flower pot fell?

Write equations for the time to reach the top and bottom of the window. There will be two unknowns: (1) unknown distance to from window top to windowsill above (h) and (2) time to reach the top of the window (t).

Solve the two equations in two unknowns.

(1/2) g t^2 = h
(1/2) g (t + 0.42)^2 = h + 1.9

h is easily eliminated by subtracting the first equation from the second.
g = 9.8 m/s^2

Solve for t fist and then use either of the two equations to compute h.

1.04 m

To find the distance between the top of the window and the windowsill, we need to consider the motion of the falling flowerpot and use the formula for the distance traveled by an object under constant acceleration.

1. First, let's find the initial velocity (u) of the flowerpot when it fell off the windowsill. Since the flowerpot was initially at rest, the initial velocity is 0 m/s.

2. Next, we need to determine the acceleration (a) of the flowerpot as it falls. The acceleration due to gravity on Earth is approximately 9.8 m/s^2, and since the flowerpot is falling downwards, the acceleration will be in the negative direction: -9.8 m/s^2.

3. Now, we can use the formula for distance (s) traveled by an object under constant acceleration:

s = ut + (1/2)at^2

where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.

In this case, we know the time (t) it takes for the pot to pass the window below, which is 0.42 s, and the height of the window (s), which is 1.90 m. We can rearrange the formula to solve for the initial position (u) as follows:

u = (2s - at^2) / (2t)

4. Plug in the known values:

u = (2 * 1.90 m - (-9.8 m/s^2) * (0.42 s)^2) / (2 * 0.42 s)

Simplifying this expression will give us the initial position (u).

5. Calculate:

u = (3.8 m - (-1.79 m)) / 0.84 s

u = 5.59 m / 0.84 s

u ≈ 6.65 m/s

6. The distance between the top of the window below and the windowsill is the initial position (u). Therefore, the distance is approximately 6.65 meters.