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Good Dog
Oct 16, 2008

Who threw this cat at me?
Clapping Larry

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. :(

I would use the x as a known (since we know the distance between the two buildings) and instead solve for y (the drop). Like if the drop was 16m instead of 15m like your assumption, that might be enough to get you over 50m traveled.

You've got gravitational acceleration down as vertical velocity.

x=x0+vx0t+.5axt2
y=y0+vy0t+.5ayt2

x is distance between buildings
x0 is 0
vx0 is how fast you want to go at takeoff
ax is 0 unless you want to get fancy and include drag (in the negative) based on the frontal area of the car
y0 is 0
vy0 is 0
ay is -9.8m/s^2

Your unknowns are t and y, and you have 2 equations to solve for them. If y is greater than however high you think the car jumps from, it'd have hit the ground before reaching the other tower.


Did these guys get enough distance to go between buildings though??

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