Pawn 17 posted:

I'm going to try to think back to HS physics class.

Distance between towers: 50m (we need to jump this far)
Drop in floors between towers: 5? (we drop ~15m)

Car jump info!!
initial horizontal velocity: 26.822 meters/second
horizontal velocity: 0 m/s
initial vertical velocity: 0 m/s
vertical velocity -9.8 m/s/s
height y: 15m (drop from 1 tower to the next)
distance traveled for the jump x: ????

using the kinematic equations we should first determine the time of flight

-15m = (0 m/s) * t + 0.5 * -9.8 m/s/s * t^2
-15m = 0.5 * -9.8 m/s/s * t^2
-15m = -4.9 m/s/s * t^2
3.06122 s^2 = t^2
total flight time t = 1.75 seconds

Now we can determine the distance traveled x

x = 26.822 m/s * 1.75 s + 0.5*(0 m/s/s) * 1.75 s ^2
total distanct traveled x = 48.47 m

SO, even if you don't account for the loss of speed breaking through the window and air resistance, the car would fall just short of the other tower and Dom would fall to a fiery death.