Velocity needed to launch a car

AI Thread Summary
The discussion revolves around calculating the velocity required for a car to launch off a roof and land 18 meters away and below, as depicted in a scene from "Transporter 2." The poster estimates the car needs to achieve a velocity of approximately 9.42 m/s (about 21 mph) to clear the distance, which seems unrealistic given the context of the scene. They express confusion over the mass of the car not affecting the velocity calculation, and acknowledge that air resistance is negligible for such a short jump. Responses confirm that 20 mph is not particularly slow, suggesting that cinematic effects may distort perceptions of speed and distance. The conversation highlights the need to consider additional factors, such as the force required to break through barriers, in their analysis.
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Homework Statement



Hi all, this is my first post here because I am having an issue with an assignment in my General Engineering class (physics related). The assignment is to find an action movie clip that portrays something to be real but is, in reality, not possible. My group and I decided on this clip from Transporter 2:



Now, we have three problems in the scene that we need to figure out. We are going to work backwards and start with the velocity a car would need to launch itself off a roof and land on something (we estimated) 18 meters across and 18 meters below. We figured the car dropped about six stories, 10 feet per story, which is approximately 18 meters. We also did some research and found a 4 lane road plus sidewalks would be about 18 meters across. We're allowed to make assumptions as obviously some things cannot be known by just watching a movie.

Homework Equations



I've done multiple kinematics equations but all of the answers seem to be too low. I guess my main question is, how can the mass of the car not be taken into consideration when trying to estimate to the velocity needed for a 4000 pound car to clear 18 meters?

The Attempt at a Solution



There are two unknowns. The initial velocity of the car, and the time it takes for the car to clear 18 meters. My attempt at solving:

-18 = Viy + (.5)(-9.8)(t2)

=

-18 = -4.9t2

3.67 = t2

t = 1.91 s

Now, knowing the time, I solve for velocity:

18 = Vi(1.91)+(.5)(0)(1.91)

=

18/1.91 = Vi

Vi = 9.42 m/s

9.42 m/s is around 21mph. I'm sitting here looking at my 6 story apartment and trying to visualize a 4000 pound car driving at that slow of pace and landing 18 meters from where it started. What am I missing here? Thanks in advance!(the force needed to break through a concrete barrier will be solved later)
 
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That calc looks good...neglecting air resistance, which should not be too much of a factor in that short time and distance. 20 mph isn't that slow...
 
True. Thanks for the reply. I guess the movie effects are just playing tricks on my eye. The kicker will have to be the force required to break a concrete barrier and sustain the velocity needed to make the jump.
 
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