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Physics Goes To The Movies! (2 Fast 2 Furious)

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by

NikRavCar Milenovic

on 16 December 2013

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Transcript of Physics Goes To The Movies! (2 Fast 2 Furious)

What Was The Initial Velocity?
The car's speed at the ramp can be calculated.
Distance?
First, the distance
the director gives
us from the shore to
the boat is:
What Are
We Testing?
Can this 1969 Camaro really jump this distance to a boat?
Angle of the Ramp?
The way we found the angle is we used an awesome
virtual protractor
over the wide
angle shot.

Approx. 17 degrees
In this video below we see the two main characters Brian and Roman attempt to catch up to a yacht in which they jump off a ramp with their car to land on it
Distance
This is the distance which the director gives us. Now using what we know is this distance
realistic?
(When compared to the car's speed).
2 Fast 2 furious
Question is.... IS IT POSSIBLE!?
Physics Goes To The Movies!
Conclusion
This movie scene contains incorrect physics!
With a car traveling at 59.2m/s it would go
approx. 200m.

In the scene the movie shows it jump a mere
30m which is off by 170m! That's quite a difference.
Sources
All data was collected from the YouTube clip and 1 Wikipedia page regarding the 1969 Camaro
bY: nIKOLA mILENOVIC / cARSON wHITLOCK / rAVEEL Tejani
Velocity 39.786 m/s
Velocity 52.7508 m/s
Time between these two is The time between these two

6 seconds 3 seconds


So What Do We Know?
Initial velocity : 59.18m/s
Distance : 30.706m
Angle : 17 degrees
Time : 3.53s
*Neglecting Air Resistance
Is This Jump Realistic?
We will use the car's initial speed, the angle of the jump and the time to calculate
the
distance
of the
actual
jump.
Calculations!
We can simplify these calculations:
- We will find
Vix
(velocity of x-direction)
and multiply it by the
time
of the jump
- With this we will find the
distance
.
Velocity ???
Vf = Vi + a(t) Vf = Vi + a(t)
52.7 = 39.7 + a(6) Vf = 52.7 + 2.16(3)
a = 2.1608m/s^2


Vf = 59.1824m/s


The car is launched at:

59.1824m/s
Is the boat too far away?
OR IS IT TOO CLOSE!?
5 Camaros long.
Using the power
of the internet we can find one
of these cars is
4.724m long.


Distance = 5 x car length
23.62m
=
5
x (
4.724m
)
BUT WAIT! Because we are the best physics 12 students ever, in order to be more accurate we added 1 1/2 Camaro's to distance because
the boat is moving and enlarging the distance.
D = 30.706m
.
Time
To find the
time
of the jump we used a simple formula
Vf
y
= Vi
y
+ a
y
(
t
)

0
=
17.308
+ (
-9.8m/s^2
)(
t
)

t
=
1.765s

This is time of the half way so,
1.765s x 2 = real time =
3.53s

vi
y
= 59.2 * sin 17 =
17.308m/s
Equation
Distance
=
Vix
*
time
Vix
= Vi * Cos 17
Vix
= 59.2m/s * Cos 17
Vix
=
56.61m/s


Time = 3.53s
Vix = 56.61m/s

Distance
=
Vix
*
time

Distance
=
56.61m/s
*
3.53s


Distance = 199.84m!!
Thanks for listening!








R.I.P Paul Walker
Full transcript