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