- #1
pikwa10
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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