5. Using the Laplace transform, we want to solve the second part of the initial value problem when the bungee jumper is 30 or more feet below the bridge. That is, we want to solve the following IVP using the Laplace Transform. me, tar2 - b(x2) = mg; for t > t r() = 0; r,(t) = v1- Since the Laplace transform requires to know the value of x2(t) at t = 0, we will define a new variable #= t -t, and a new function y2() = ry( + 1). Notice that this is just applying a horizontal shift to 12, which will not change it's derivatives. Thus yz would satisfy the same differential equation, but have the following initial conditions, my, + ayz + kym = mg; 32(0) = =2(t1) = 0; 1/2(0) = 12(t1) = 01. We will solve this shifted initial value problem for y2() using the Laplace transform, then apply 12(/) = x2(u + ti) = x2(t). Again, you may use a = 2.8 and g = 9.8, but leave m and & as unknown constants. The solution 2(t) represents your position below the natural length of the cord after it starts to pull back. (I recommend that you leave v1, a, and g as variables when find the solution to the IVP, and only substitute the values of these three variables at the end.)