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A ball dropped from the top of the building takes 0.2 s to pass a window 4.2 m high. How far is the top of the window below the top of the building?

Question

A ball dropped from the top of the building takes 0.2 s to pass a window 4.2 m high. How far is the top of the window below the top of the building?

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Solution

To solve this problem, we need to use the equation of motion which is:

d = ut + 0.5gt^2

where: d = distance u = initial velocity t = time g = acceleration due to gravity

Given in the problem: u = 0 (since the ball is dropped, not thrown) t = 0.2 s (time taken to pass the window) g = 9.8 m/s^2 (approximate value of acceleration due to gravity) d = ? (we need to find this)

Substituting the given values into the equation:

d = 0*(0.2) + 0.59.8(0.2)^2 d = 0 + 0.59.80.04 d = 0.196 m

This is the distance the ball falls in the time it takes to pass the window. However, we know that the window is 4.2 m high, so the ball must have fallen from a point above the window. Therefore, the top of the window is:

4.2 m (height of the window) + 0.196 m (distance the ball

This problem has been solved

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