Velocity needed to launch a car

Click For Summary
SUMMARY

The discussion centers on calculating the velocity required for a 4000-pound car to launch off a roof and clear a distance of 18 meters horizontally while descending 18 meters vertically. The calculations indicate that an initial velocity of approximately 9.42 m/s (21 mph) is necessary, derived from kinematic equations. The participants acknowledge that the mass of the car does not directly affect the velocity needed for the jump, as the calculations focus on the time of flight and horizontal distance. The impact of air resistance is deemed negligible for this scenario, and the discussion highlights the unrealistic portrayal of such stunts in action movies.

PREREQUISITES
  • Understanding of kinematics and projectile motion
  • Familiarity with basic physics equations, specifically those related to motion
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Ability to perform unit conversions and basic algebra
NEXT STEPS
  • Research advanced kinematics involving air resistance and its effects on projectile motion
  • Explore the physics of impacts, specifically the force required to break through barriers
  • Study real-world applications of projectile motion in automotive engineering
  • Investigate the effects of mass on different types of motion and energy transfer
USEFUL FOR

Students in physics or engineering courses, automotive engineers, and anyone interested in the realistic application of physics in action films.

NEUhusky
Messages
5
Reaction score
0

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)
 
Last edited by a moderator:
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
4K
Replies
10
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
2K
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 2 ·
Replies
2
Views
10K
Replies
16
Views
2K