# Answer on Mechanics | Relativity Question 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.

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>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>0 )

y = 21.168

Therefore:

h = y^2 = 209.542 m

Answer: h = 209.54 m

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>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>0 )

y = 21.168

Therefore:

h = y^2 = 209.542 m

Answer: h = 209.54 m

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