Ok, let's try some calcs.. I think I'll try meters/second. So, above I calculated 69 feet, or 21 meters.

v = velocity leaving bridge

v_{h} = horizontal velocity (constant) leaving the bridge

v_{vi} = initial vertical velocity leaving the bridge

d_{v} = vertical distance relative to the top of the ramp.

t = time for the jump.

a = the acceleration due to gravity (9.8[tex]\frac{m}{s^2}[/tex])

Where

[tex]\frac{v}{\sqrt{2}}[/tex] = v_{h} = v_{vi}

t = [tex]\frac{21m}{v_h}[/tex] = [tex]\frac{21m}{v_{vi}}[/tex]

For the vertical component of the velocity and distance:

v_{v} = v_{vi} - at

d_{v} = v_{vi} t - ½ a t^{2}

The midpoint in the flight would be setting v_{v} = 0

So, substituting half the time above:

v_{v} = 0 = v_{vi} - 9.8 * ½ * [tex]\frac{21m}{v_{vi}}[/tex]

0 = [tex]\frac{v_{vi}^2}{v_{vi}}[/tex] - [tex]\frac{102.9}{v_{vi}}[/tex]

v_{vi} = [tex]\sqrt{102.9)[/tex]m/s = 10.14 m/s

Set d_{v} = 0 (landing), and substitute in time from above, and one gets:

d_{v} = 0 = v_{vi} * [tex]\frac{21m}{v_{vi}}[/tex] - ½ * 9.8 * [tex]({\frac{21m}{v_{vi}}})^2[/tex]

0 = [tex]21m*\frac{v_{vi}^2}{v_{vi}^2}[/tex] - 21m* 4.9 * [tex]\frac{21m}{v_{vi}^2}[/tex]

0 = v_{vi}^{2} - 102.9

v_{vi} = [tex]\sqrt{102.9)[/tex]m/s = 10.14 m/s Whew, the same as above when calculating with velocity.

t = [tex]\frac{21m}{10.14 m/s}[/tex] = 2.07 seconds.

Now, to recover the initial velocity, v = 10.14 m/s * [tex]\sqrt{2}[/tex] = 14.3 m/s = 32 mph

That is actually a lot slower than I had expected, but it makes sense for about 2 seconds of air time.

On the level, a good car should be able to reach 32 mph in a little less than 1/8 mile, or about 200 meters. Perhaps a lot less as the low speed acceleration is always easiest.

So, from your drawing, assuming a safe, gentle transition from horizontal to the 45° angle, one may be able to do the run, starting just off of the bridge, just before the first tower.