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# Mechanics

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A steel ball is dropped from a building's roof and passes a window, taking 0.125 s  to fall from the top to the bottom of the window, a distance of 1.20m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.125 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 2.0 s. How tall is the building?

Kreyn  Mar 5, 2017
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#1
+12117
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I get an answer of 10.32 m   .....anyone else get the same?

ElectricPavlov  Mar 5, 2017
#2
+6531
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I don't really know how to do this but I was just thinking about it and if we just knew the total time of the fall then we could figure out how high it fell from because we know gravity makes stuff fall at about 9.8 m/s2 . (But you have to forget about air resistance.)

So wouldn't the time of the fall be 2.0 + 0.125 + 0.125? Because if the bounce up was identical to the fall then there is exactly as much height above the window as below it.

So the total time is 2.25 seconds...and gravity makes stuff fall at 9.8 m/s/s.

So height divided by time divided by time is 9.8

height divided by 2.25 divided by 2.25 is 9.8

height is 9.8*2.25*2.25

That's probably wrong though because I'm just guessing at how to do it.

hectictar  Mar 5, 2017
#3
+12117
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First we will find the distance to the TOP of the window:

x=xo + vot + 1/2 at^2      Where xo = 0 (top of window)  and x = window height = 1.2 m

1.2 = vo (.125) + 1/2 (9.8)(.125^2)

vo= 8.9875 m/s   at top of window

NOW:  v = vo + at     where  vo = 0 (at start)

8.9875 = 0 + 9.8 (t)

t = time to reach window top = .91709 seconds

NOW  Distance to TOP of window :

x = xo + vot + 1/2 a t^2

= 0   + (0)t + 1/2 (9.8) (.91709)^2

x = 4.12 m  to top of window

NOW it takes 1 second + .125 second to travel back to the top of the window = 1.125 seconds

again:  x = xo + vot + 1/2 at^2

= 0 + 0(t) + 1/2(9.8)(1.125^2) = 6.20 meters

Finally:

6.20 meters + 4.12 meters = 10.32 meters tall

ElectricPavlov  Mar 5, 2017
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+104
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