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# Answer to Question #26135 in Mechanics | Relativity for Jake Thomson

Question #26135
Spiderman steps from the top of a tall building. He falls freely from rest to the ground a distance of h. He falls a distance of h/ 3 in the last interval of time of 1.2s of his fall.

What is the height h, in metres, of the building? Express your answer using two significant figures.
1
2013-03-12T09:59:08-0400
After the spiderman falls 2/3h:
m*v^2/2 = m*g*2/3*h & -& energy conservation law
v=Sqrt[ 4g*h/3 ]
He falls freely, so its acceleration equals g and for last h/3 we have:
h/3 = v*t + g*t^2/2
Substitute v=Sqrt[ 4g*h/3 ] :
h/3 = Sqrt[ 4g*h/3 ]*t + g*t^2/2
Sqrt[h]=y , y&gt;0:
y^2 - 2t*Sqrt[3g]*y - 3g*t^2/2 = 0
y^2 - 13.0132*y - 21.168 = 0
Solving:
y = -1.46233 ( but in our case y&gt;0 )
y = 21.168
Therefore:
h = y^2 = 209.542 m

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