Posted by monnie on Friday, October 8, 2010 at 6:00pm.
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 2.13 s. How tall is the building?

Physics  bobpursley, Friday, October 8, 2010 at 6:44pm
figure the average velocity in the window (distance/time). That is the velocity in the center of the window. call that Va
Va^2=2gd solve for d, the distance from the top of the building to the center of the window.
Now do the same for the upward flight.
Notice it took 2.13 sec/2 to hit the ground, so
hf=Va*1.065 4.9*1.065^2 solve for hf, the distance below the center of the window.
How tall is the building? hf+d
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