Ignoring air resistance (as this has to do with gravity), it should be as simple as using a vector equation.

Let's say $Sx(t)$ is horizontal position (for how far one could jump), and $Sy(t)$ is vertical position (for how high one could jump). Then it's as simple as this:

1. Determine gravity $(g)$ in distance units per time units squared. (Often meters per second squared, or feet per second squared)
2. Determine initial horizontal velocity $(Vx0)$ and initial vertical velocity $(Vy0)$. This is how fast you're going when you first jump, which is determined by lots of factors in muscles and such.
3. Set $Sx(t) = Vx0 * t + Sx0$, and$Sy(t) = -(g/2) * t^2 + Vy0 * t + Sy0$, where S0 is starting hieght. (If starting on the ground, you can often set S0 to 0)
4. Combine the functions into a vector function, with the points $(Sx(t), Sy(t))$.

This can be done easily in desmos.com/calculator, where you can easily type the x function and y function, and then put them together as an "ordered pair" (vector function) to see how high and far you'd jump.

Jumping at a 45 degree angle results in maximum distance, which simply means that $Vx0 = Vy0$

Ignoring air resistance (as this has to do with gravity), it should be as simple as using a vector equation.

Let's say $Sx(t)$ is horizontal position (for how far one could jump), and $Sy(t)$ is vertical position (for how high one could jump). Then it's as simple as this:

1. Determine gravity $(g)$ in distance units per time units squared. (Often meters per second squared, or feet per second squared)
2. Determine initial horizontal velocity $(Vx0)$ and initial vertical velocity $(Vy0)$. This is how fast you're going when you first jump, which is determined by lots of factors in muscles and such.
3. Set $Sx(t) = Vx0 * t + Sx0$, and$Sy(t) = -(g/2) * t^2 + Vy0 * t + Sy0$, where S0 is starting hieght. (If starting on the ground, you can often set S0 to 0)
4. Combine the functions into a vector function, with the points $(Sx(t), Sy(t))$.

This can be done easily in desmos.com/calculator, where you can easily type the x function and y function, and then put them together as an "ordered pair" (vector function) to see how high and far you'd jump.

Jumping at a 45 degree angle results in maximum distance, which simply means that $Vx0 = Vy0$

In short, if you halve gravity, but initial velocity is the same (not guaranteed), you can jump double the distance. But you have to consider how you body works when jumping, and how gravity affects it. Lower gravity would make it more difficult to stay on the surface and push off with the same force, so ir may be less than double. From a strictly numeral basis, though, it can be easily determined with the above equations.