February 21st, 2007 @ 11:15CT by kangsta
Commercial text not included: “Based on horizontal drop. Aerial sequence simulated. Professional driver on closed course. Do not attempt.”
Wiki’ing + Googling + recalling basic HS physics yields:
Terminal Velocity: V = sqrt ( (2 * W) / (Cd * r * A)
v=terminal velocity
W=weight in Newtons
Cd=Drag Coefficeint
r=atmospheric density
a=area exposed to air(area facing down)
^ simplified aerodynamics fluid formula
2006 Lexus IS350 is 3,435 lbs
Atmospheric Density (hotter) = 1.184
Assuming the car is falling flat - dimensions are 180in x 70.9 in exposed surface (88.625 sqr ft)
Drag of Lexus nose-forward is 0.3 falling cube is 0.8 … so let’s say 0.6
Thus…
3000 ft 243.795 ft/s
2000 ft 240.198 ft/s
1000 ft 236.679 ft/s
0 ft 233.235 ft/s
We can disregard the first 1000ft since it won’t hit terminal velocity. So…
V=at thus
We reach terminal velocity (240ft/s) with gravity 32ft/s in 7.456s.
d=1/2at
Thus, the car travels 894.688ft before hitting terminal velocity. Considering the distance travelled, and now hitting termianl velocity…
the car has another 3105ft or 12.939s left to fall.
Thus… the car takes about ~20s to fall to the ground. No way to get that drag coefficent without a physics lab.
A Lexus takes 19.231s to travel 4000ft. (Max speed of a Lexus 2006 IS350 is 140mph)
BUT this does not take into account the drag, wind resistance, and acceleration of the Lexus. Basically, no way in hell would that car on the ground make it in time.
Lexus: FAILS
INC False advertisement suits?