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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.
Expert's answer
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
Answer: h = 209.54 m

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