25
Depends on your launch angle. Acceleration due to gravity is constant.
45 degrees is the theoretical ideal launch angle to maximize horizontal distance, but accounting for wind resistance make it closer to 42 degrees. Since you won't be doing it for real, let's say 45 degrees to keep the math easy. In fact, we're ignoring air resistance and friction of all kinds. If you want to get real, use a glider.
At its narrowest point, Marble Canyon, it's about 600 feet across. It could be as much as 18 miles, so let's start small and go from there.
S is speed. Vx is horizontal velocity, Vy is vertical velocity, and t is time in the air. X is the distance across the canyon. Y is only necessary if the two sides have different elevations, but let's ignore that, too.
The time in the air is how long it takes for gravity to make the vertical velocity -Vy.
X = Vxt
0 = Vy + gt/2 so -gt/2 = Vy and t = -2Vy/g
S^2 = Vx^2 + Vy^2 and at 45 degrees, Vx = Vy so S = (√2)Vx
Replace some terms, and we get
X = Vx(-2Vx/g) so X = -2(Vx)^2/g
√(-Xg/2) = Vx and S = (√2)(√-Xg/2)
S = √(-Xg)
So if X is 183 meters at the smallest, and g is -9.8 meters per second squared, then you need a speed of 42.35 m/s at launch, or just shy of 95 mph. You will be in the air for about 4.3 seconds. That's theoretically possible, but remeber you'd be landing while traveling at close to 95 mph at a 45 degree angle towards the ground. The jump is just half the battle.
If you go to the average width, 10 miles or 16,000 meters, requiring a speed of 397 m/s, or 888 mph. At its widest 18 miles, 29,000 meters, you need a speed of 533 m/s or 1,192 mph. At that speed, it's a good thing we're ignoring friction, because air resistance would start to make things toasty.