A falling stone takes 0.33 s to travel past a window 22m tall. From what height above the top of the window did the stone fall?

To find the height from which the stone fell, we can use the equations of motion under gravity. Given that the stone takes 0.33 seconds to pass a window that is 22 meters tall, we can first calculate its speed as it passes the window.

Using the formula for distance traveled under constant acceleration (in this case, gravity):

d = v * t + (1/2) * g * t^2

Here:

• d = distance traveled (22 m)
• t = time taken to travel past the window (0.33 s)
• g = acceleration due to gravity (approximately 9.81 m/s²)
• v = initial velocity when it enters the window (which we need to find)

Rearranging it, we can derive an expression for v:

22 = v * 0.33 + (1/2) * 9.81 * (0.33)^2

Calculating the second term:

(1/2) * 9.81 * (0.33)^2 ≈ 0.538 m

So we have:

22 = v * 0.33 + 0.538

Now solving for v:

v * 0.33 = 22 - 0.538

v * 0.33 = 21.462

v ≈ 65.12 m/s

This speed is the velocity of the stone just before it enters the window. Now we want to find the height from which the stone fell, starting from rest and falling to the point just above the window:

We can use another equation of motion:

v² = u² + 2as

Where:

• u = initial velocity (0 m/s, since it starts from rest)
• a = acceleration (9.81 m/s²)
• s = the distance fallen from the height above the window to the point just above the window.
• v = 65.12 m/s (from our previous calculation)

Substituting the known values:

(65.12)² = 0 + 2 * 9.81 * s

4255.6544 = 19.62 * s

s ≈ 216.55 m

Therefore, the stone fell from a height of approximately 216.55 meters above the top of the window. This shows us not only how fast the stone was traveling but also how far it fell before reaching the window.