Let's analyze the forces acting on an object that falls onto a trampoline:
Suppose we have a trampoline, of width w, with elastic properties similar to a spring of spring constant k with its left edge situated at a fixed point, A, and its right edge situated at a fixed point, B. An object lands on the trampoline at a horizontal distance x from the left edge at an initial height, hi and depresses the trampoline vertically to a final height, hf. If we take the distance depressed to be infinitesimally small, that is Δh→0, then we can obtain an accurate model for the forces in action right when the object hits the trampoline. If rest position is taken to be when the trampoline is horizontal, then the rest length of the left spring is x and the rest length of the right spring is w-x.
After the point on the trampoline where the object lands is depressed a displacement Δh, the left spring now forms the hypotenuse of a triangle with an adjacent (surface) length of x and an opposite (depth) length of Δh, so its new length is \sqrt{}(x2+(Δh)2). Likewise, the right spring now forms the hypotenuse of a triangle with an adjacent (surface) length of w-x and an opposite (depth) length of dh, so its new length is \sqrt{}((w-x)2+(Δh)2).
In order to find the forces these springs exert, notice that each spring undergoes a stretch, Δl equal to the difference of its stretched- and rest-state lengths:
The left spring undegoes a stretch, Δll =\sqrt{}(x2+(Δh)2)-x.
The right spring undergoes a stretch, Δlr =\sqrt{}((w-x)2+(Δh)2)-(w-x).
Now, applying Hooke's Law, F=-kΔl, and that both springs have identical spring constants:
The left spring exerts an elastic pull force on the object, Fl, equal to kΔll along its stretched axis away from the object and toward A.
Similarly, the right spring exerts an elastic pull force on the object, Fr, equal to kΔlr along its stretched axis away from the object and toward B.
Now examine the forces acting on the object by breaking them down into x- and y-components:
To get the x- and y-components of these forces, consider that the force vector of a spring, F, is aligned along the spring in its stretched state, which is the resultant (i.e. the hypotenuse) of a force triangle with horizontal and vertical components, so a ratio of sidelengths can be used to find the x- and y-components of both force vectors.
The x-component forces acting on the object are the x-components of the two elastic pull forces:
Fxl = Fl X x /\sqrt{}(x2+(Δh)2)
Fxr = Fr X (w-x) /\sqrt{}((w-x)2+(Δh)2)
The sum of the x-component forces acting on the object, ∑Fx, equals the force acting to the right minus the force acting to the left,
Fxr-Fxl = Fr X (w-x) /\sqrt{}((w-x)2+(Δh)2) - Fl X x /\sqrt{}(x2+(Δh)2)
The y-component forces acting on the object are the y-components of the two elastic pull forces and gravity:
Fyl = Fl X Δh /\sqrt{}(x2+(Δh)2)
Fyr = Fr X Δh /\sqrt{}((w-x)2+(Δh)2)
Wobject =Mobject X g
The sum of the y-component forces acting on the object, ∑Fy, equals the forces acting up minus the force acting down,
Fyl + Fyr - Wobject = Fl X Δh /\sqrt{}(x2+(Δh)2) + Fr X Δh /\sqrt{}((w-x)2+(Δh)2) - Mobject X g
Let's say your trampoline is 5 m wide, and its bed can be represented by springs with k=2000 N/m. You push down vertically 1 m from the right edge of the bed, depressing the surface .2 m. The left spring has gone from a rest length of 4 m to a stretched length of
\sqrt{}((4 m)2 + (.2 m)2) ≈ 4.005 m, so it has stretched approximately .005 m, and it now exerts an elastic pull force on your hand approximately equal to .005 m X 2000 N/m = 10 N toward the point where the left edge of the bed is attached to the trampoline frame. The right spring has gone from a rest length of 1 m to a stretched length of \sqrt{}((1 m)2 + (.2 m)2) ≈ 1.020 m, so it has stretched approximately .020 m, and it now exerts an elastic pull force on your hand approximately equal to .020 m X 2000 N/m = 40 N toward the point where the right edge of the bed is attached to the trampoline frame. Both of these force vectors point diagonally up and away from each other. The ratio of the x-component of the left force vector to its resultant is 4/\sqrt{}(42+(.02)2) ≈ 1, and the ratio of the x-component of the right force vector to its resultant is 1/\sqrt{}(12+(.02)2) ≈ 1, so the x-components of these vectors are pretty much equal to the values of the vectors themselves. Therefore, the right spring pulls with approximately 40 N of force to the right, and the left spring pulls with approximately 10 N of force to the left, so the net horizontal force acting on your hand is approximately 30 N to the right. The vertical components of the elastic pull forces both pull up on your hand.
The ratio of the y-component of the left force vector to its resultant is .2/\sqrt{}(42+(.02)2) ≈ .050, and the ratio of the y-component of the right force vector to its resultant is .2/\sqrt{}(12+(.02)2) ≈ .200. So the left spring exerts a vertical force of 10 N X .050 ≈ .5 N, and the right spring exerts a vertical force of 40 N X .200 ≈ 8 N, so in order to keep your hand where it is, you must push down with approximately 8.5 N of force and to the left with approximately 30 N of force.
However, an object that falls onto a trampoline cannot push to the left or to the right, so its horizontal motion depends entirely on how the "springs" behave. So if you fall onto the bed of a trampoline closer to one edge of the bed, your horizontal motion will be in the direction of that edge as you depress the bed, and once you reach your minimum height, the bed will rebound, pulling you back up and away from the edge and toward the center, so the answer to your question is that you will get launched back toward the center, though this effect probably won't be very noticeable on a real trampoline unless the bed is highly elastic (like a spring). Even then, the effect will be very minimal since most jumpers will not displace the bed surface anywhere near .2 m; however, these parameters are up to you. Have fun with your simulation.
