What Is the Height of the Building in This Free Fall Acceleration Problem?

AI Thread Summary
The discussion centers on a physics problem involving a steel ball dropped from a building, with specific parameters given for its fall and bounce past a window. The user attempts to calculate the height of the building using kinematic equations but arrives at an incorrect conclusion. Key calculations include the time taken for the ball to pass the window and the total time of fall, which are crucial for determining the building's height. The user is advised that their mistake lies in the application of the equations and the interpretation of the time intervals. Correcting these errors is essential for accurately solving the problem and finding the building's height.
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Homework Statement



Hello. Can someone please give me a clue as to what I'm doing wrong and why I'm doing it wrong? Thank you. The problem is:

A steel ball is dropped from a building's roof and passes a window, taking 0.13 s to fall from the top to the bottom of the window, a distance of 1.20 m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.13 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is 2.23 s. How tall is the building?

Homework Equations



V= Vo + at
X - Xo= Vo t + 1/2 a t^2

The Attempt at a Solution



V= -1.20 m/ 0.13 s = - 9.23 m/s (velocity of ball when it just passed the window)

-9.23 m/s= 0 m/s + (-9.8 m/s^2) t ---> t= 0.94 s to reach - 9.23 m/s

0.94 s to reach the window + 0.13 s to pass the window + 2.23/2 (1.12 s) to hit the ground= t total= 2.19 s to reach sidewalk

delta x= 0 m/s (2.19 s) + 1/2 (-0.9 m/s^2) (2.19 s) ^2

delta x= -24 m ---> 24 m is roughly the height of the building
 
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V= Vo + at

V= -1.20 m/ 0.13 s = - 9.23 m/s (velocity of ball when it just passed the window)


Here's your mistake.
 
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