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# A first course in differential equations bungee jumping problem

#### Bydreamtravel

Sep 7, 2023

### A First Course in Differential Equations: The Bungee Jumping Problem
In this article, we will use a first-order differential equation to model the motion of a bungee jumper and solve it to find the height of the jumper as a function of time.

#### The Model
Let (h(t)) be the height of the bungee jumper at time (t). The velocity of the jumper is given by (v(t) = frac{dh}{dt}). The acceleration of the jumper is given by (a(t) = frac{dv}{dt}).

The forces acting on the jumper are gravity and the force of the bungee cord. The force of gravity is given by (F_g = mg), where (m) is the mass of the jumper and (g) is the acceleration due to gravity. The force of the bungee cord is given by (F_c = -kx), where (k) is the spring constant of the bungee cord and (x) is the displacement of the bungee cord from its equilibrium length.

The net force acting on the jumper is the sum of the force of gravity and the force of the bungee cord:

\$\$F_{net} = F_g + F_c = mg – kx\$\$

Newton’s second law states that the net force acting on an object is equal to the mass of the object times its acceleration:

\$\$F_{net} = ma\$\$

Substituting (F_{net}), (m), and (a) into Newton’s second law, we get:

\$\$mg – kx = mfrac{dv}{dt}\$\$

### Solution to the Differential Equation:
>

\$\$mg – kx = mfrac{dv}{dt}\$\$
\$\$mfrac{dv}{dt}= mg – kx\$\$
\$\$frac{dv}{dt}=g – frac{k}{m}x\$\$
\$\$int frac{dv}{dt} dt =int (g – frac{k}{m}x) dt\$\$
\$\$v=gt – frac{k}{2m}x^2\$\$
\$\$frac{dx}{dt}=gt – frac{k}{2m}x^2\$\$
\$\$(gt – frac{k}{2m}x^2)^{-1}frac{dx}{dt}=1\$\$
\$\$int (gt – frac{k}{2m}x^2)^{-1}frac{dx}{dt} dt= int 1 dt\$\$
\$\$frac{2m}{k}arctg(frac{xsqrt{frac{k}{2m}}}{gt}) = t+C\$\$
\$\$x = sqrt{frac{2mg}{k}}gt tan (frac{k}{2mg} (t+C))\$\$

#### Using the Initial Conditions
At (t = 0), the jumper is at the top of the jump, so (h(0) = H). Substituting (t = 0) and (h(0) = H) into the expression for (x), we get:

\$\$sqrt{frac{2mg}{k}}g(0)tan (frac{k}{2mg} (0+C)) = H\$\$

>Therefore,
\$\$C= arctan(frac{H}{sqrt{frac{2mg}{k}}g})\$\$

#### Result
Substituting (C) into the expression for (x), we get:

\$\$x = sqrt{frac{2mg}{k}}gt tan (frac{k}{2mg} (t+arctan(frac{H}{sqrt{frac{2mg}{k}}g})))\$\$

\$\$h(t) = H-x = sqrt{frac{2mg}{k}}g t tan (frac{k}{2mg} (t+arctan(frac{H}{sqrt{frac{2mg}{k}}g})))\$\$

This is the general solution to the differential equation.

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