Comments on: A roofing tile falls from rest off the roof of a building.? http://www.allroofingmaterials.com/tile-roofing/a-roofing-tile-falls-from-rest-off-the-roof-of-a-building Materials for your roof Thu, 01 Apr 2010 17:57:59 +0000 hourly 1 http://wordpress.org/?v=3.0 By: Physicsquest http://www.allroofingmaterials.com/tile-roofing/a-roofing-tile-falls-from-rest-off-the-roof-of-a-building/comment-page-1#comment-2035 Physicsquest Thu, 04 Mar 2010 21:06:59 +0000 http://www.allroofingmaterials.com/tile-roofing/a-roofing-tile-falls-from-rest-off-the-roof-of-a-building#comment-2035 2.2/0.37 = average speed = 5.946 m/s velocity at top sill = 5.946 - ((0.37/2) x 9.8) = 4.133 m/s v = at 4.133/9.8 = t = 0.42173 secs since the tile fell from rest s = 1/2gt^2 4.9 x 0.42173^2 = 0.8715 m (answer)<br><b>References : </b><br> 2.2/0.37 = average speed = 5.946 m/s

velocity at top sill = 5.946 – ((0.37/2) x 9.8) = 4.133 m/s

v = at

4.133/9.8 = t = 0.42173 secs

since the tile fell from rest

s = 1/2gt^2

4.9 x 0.42173^2 = 0.8715 m (answer)
References :

]]> By: RickB http://www.allroofingmaterials.com/tile-roofing/a-roofing-tile-falls-from-rest-off-the-roof-of-a-building/comment-page-1#comment-2034 RickB Thu, 04 Mar 2010 20:17:59 +0000 http://www.allroofingmaterials.com/tile-roofing/a-roofing-tile-falls-from-rest-off-the-roof-of-a-building#comment-2034 The distance the tile falls in a given time "t" is: d = ½gt² Apply this to the the upper windowsill: call its distance (from the roof) "d1", and the time "t1": d1 = ½g(t1)² Apply it to lower windowsill ("d2" and "t2"): d2 = ½g(t2)² > "...windowsills that are 2.2 m apart..." This means: d2 - d1 = 2.2m or: ½g(t2)² - ½g(t1)² = 2.2m or, simplified: ½g((t2)² - (t1)²) = 2.2m > "...takes 0.37 s " This means: t2 - t1 = 0.37 s The rest is algebra. Solve these simultaneous equations to get t1 and t2: ½g((t2)² - (t1)²) = 2.2m t2 - t1 = 0.37 s Then, use "d1 = ½g(t1)²" to get d1<br><b>References : </b><br> The distance the tile falls in a given time "t" is:

d = ½gt²

Apply this to the the upper windowsill: call its distance (from the roof) "d1", and the time "t1":
d1 = ½g(t1)²

Apply it to lower windowsill ("d2" and "t2"):
d2 = ½g(t2)²

> "…windowsills that are 2.2 m apart…"
This means:
d2 – d1 = 2.2m
or:
½g(t2)² – ½g(t1)² = 2.2m
or, simplified:
½g((t2)² – (t1)²) = 2.2m

> "…takes 0.37 s "
This means:
t2 – t1 = 0.37 s

The rest is algebra. Solve these simultaneous equations to get t1 and t2:
½g((t2)² – (t1)²) = 2.2m
t2 – t1 = 0.37 s

Then, use "d1 = ½g(t1)²" to get d1
References :

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