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 $Vx0 = Vy0$