The Human Slingshot

By Adam Wheeler

The situation I decided to do my physics news on is the Human Sling Shot.
A person is harnessed into the middle of two bungees and secured so that she does
not actually leave the harness when it is pulled back. Then an ATV hooks up to her
via the back of the harness and then takes off in the opposite direction until the
bungees are stretched significantly. Then it is released and the woman begins to fly
forward with a great amount of speed and recoils backwards until coming to a stop.
I decided to try and determine the distance that she would have gone if she was not
connected to the bungee and could actually fire out of the harness.

http://www.youtube.com/watch?v=u2-od4n5Xl0

KNOWNS:

Mass of the woman: 60 kg
K spring constant: 53 N M
Height of the poles connected to the bungee: 30 meters
θ of the bungee with regards to the pole: 20°
woman is 1.5 meters from the ground
she is shot an an angle of 45° with respects to the horizontal
the bungee length is 31m
we assume the bungee is massless

First, I found the length of the bungee cord when she was just hanging there and not
being stretched back.

Cos(20°)= 29m/X

X=29m/Cos(20°)
X=31m

Next I found the PE of the system of the girl plus the two bungees:

PE= 2(1/2* 53N*M * (30m)2) = 48,000 Joules

Therefore,

PE1 = KE2
48,000= .5(60)v2
40m/s= v

Next, we can take this velocity and say that

KE1 = KE2 + ΔPE

48,000= (60)9.8m/s2(-1.5) + .5(60)(vf)2
49,000=.5(60)(vf)2
40.4 = v

Then we find the time it takes the person to land in the y direction

Δy= V0t +.5at2

-1.5= 28t + -4.9t2

-4.9t2 + 28t +1.5=0

Quadratic formula gives: 5.8 seconds

Therefore we can find ΔX from ΔX= X0 + v0t +.5at2

ΔX= V0t
ΔX= Cos(45°)40(5.8)
ΔX=160m

The force the ground exerts on her when she hits the ground is:

F=Δp/Δt. If it takes her .5 seconds to stop then the force is equal to:
F=60(-40.4)/.5
F= -5000Newtons.