By Kody Lyng

This past week, a new trailer for the upcoming sequel to

*The Amazing Spider Man*series was released. In one of the clips, Spider-Man falls from the top of the Empire State Building in order to pursue a criminal. I would like to take a closer look at the tension force in Spider-Man’s web when he is at the bottom of his fall. We must first make a few assumptions:
Spider-Man shoots his web at the top of the Empire State
Building

At the bottom of his fall, he is 1 meter above the ground

His web does not stretch at all

Spider-Man weighs 62 kg

Ignore air resistance

In order to find his velocity at the bottom of his swing, we
must set his kinetic energy at the bottom of the swing equal to his initial
potential energy.

1/2mv

^{2}= mgh
v = sqrt(2gh) = sqrt[2*9.8 m/s

^{2}*(443m-1m)] = 93.1 m/s
In order to find the tension force of the web, we must set
the net force equal to the tension force subtracted by the force of gravity.
The net force will be the centripetal force or

F

_{c}= m*v^{2}/r
So…

m*v

^{2}/r = F_{T}– mg or F_{T}= mv^{2}/r + mg
F

_{T}= (62 kg)*(93.1 m/s)^{2}/(443 m-1 m) + (62 kg)*(9.8 m/s^{2})
F

_{T}= 1820 Newtons
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