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
)
Where
= v
_{h} = v
_{vi} t =
=
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 *
0 =

v
_{vi} =
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} *
 ½ * 9.8 *
0 =
 21m* 4.9 *
0 = v
_{vi}^{2}  102.9
v
_{vi} =
m/s = 10.14 m/s Whew, the same as above when calculating with velocity.
t =
= 2.07 seconds.
Now, to recover the initial velocity, v = 10.14 m/s *
= 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.