Q:

Find force required for a launch between two points

Let me start by saying this is within a game environment, so gravity isn't 9.81m/s^2, and the unit of measure for distance will be "blocks".
I'm attempting to find the amount of force needed in order to launch a player into the air, from point A to point B. For my example, I assumed that it was in a straight line of 60 blocks away, and attempted to work from there. Knowing that velocity is the derivative of distance, and acceleration the derivative of velocity, I mustered together those variables, along with others.
In the game:

Gravity pulls at 13 blocks/second
Acceleration to the ground is at 22.48 blocks/second^2
A block is a distance of 1 meter.

I have a general mockup of attempting to start the problem, but to be honest I'm completely stumped on where to take it. I need to find the initial velocity to launch at, as well as what angle to do it at.
General mockup: http://i.imgur.com/zMQuAbW.jpg

A:

The force isn't constant, so that approach won't get you far. Acceleration to the ground, however is constant. What you need to calculate is just the initial velocity that will keep you in the air long enough to propel you to your target. This is just a classic projectile motion problem.
If $g$ is the acceleration due to gravity -- in whatever units -- you have that your acceleration downward is $g$. Then, your height (assuming you start at the ground $h=0$) as a function of time is $h=v_y t -\frac{1}{2}g t^2$ where $v_y$ is your velocity vertically. Next, you have your distance forward, $x$, given by $x = v_x t - \frac{1}{2} a t^2$ where $v_x$ is your velocity horizontally and $a$ is your horizontal acceleration. (I'm not sure from the problem statement if that's what was meant by 'gravity pulls' or not.) So set $x$ to 60, $a$ to whatever $a$ should be (0?) and solve for $t$. Then, plug this back into the equation for $h$ and solve for $v_y$. Since this is a game, you can presumably play with values of $t$ and $v_x$ to get the results you want.
If you know the angle of the jump, $v_x = v_{total} \cos(\theta)$ and $v_y = v_{total} \sin(\theta)$ where $\theta$ is the angle from the horizontal.
If you solve these two equations you'll have your answer for your initial and final velocities. If, however, you need more than that, meaning the actual force that has to be exerted while in contact with the ground, you probably want to consider the impulse. This would be the change in momentum. (mass)(change in velocity). Dividing this by the time in contact with the ground gives you an average force that would change your momentum accordingly. Also, for force $F=\frac{d p}{dt} \sim \frac{\Delta p}{\Delta t}$ where $p$ is your momentum.

