Mythbusters Car Drop

1. Jan 11, 2010

danielatha4

Car A is dropped at 4000 ft where directly below is an x on the road. At the instant Car A is dropped Car B is exactly 4000 ft from the x driving down the road at a constant velocity of 142 mph. Does Car B make it past the x in time to avoid disaster?

It's such a simple kinematics problem that it's beautiful but unfortunately you have to take into consideration wind resistance Can anyone show an accurate application of wind resistance in this problem? From my understanding the mass of the car is 1600 kg and it dropped staying flat landing virtually on all 4 tires at the same time.

For the heck of it I'll do the kinematics neglecting wind resistance:

time Car B takes to get to the x

x=vt
1219.2m = 63.5 m/s * t
t = 19.2 seconds

Does Car A hit the road is less time or more time than 19.2 seconds?

x=vot+1/2at2
x = 0 + 4.9 * 19.22
x = 1806.336m

therefore, less time

time it takes Car A to hit the road

x = 1/2at2
1219.2m = 4.9 * t2
t = 15.77 seconds

Seems like air resistance and terminal velocity make all the difference here

2. Jan 11, 2010

Staff: Mentor

142 mph would probably be pretty close to the terminal velocity of a car (maybe even above), so I expect the driving car would get there first.

3. Jan 11, 2010

mgb_phys

Terminal velocity is sqrt( 2mg / density A Cd )

BMW 3 series = 4.4m * 1.7m 1465kg
Assuming that on four wheels Cd = 1
And air is density = 1.2kg/m^3

terminal V = sqrt ( (2 * 1465 * 9.8) / (1.2 * 4.4 * 1.7 * 1)) = 56.5 m/s = 126mph

4. Jan 11, 2010

danielatha4

Does anyone know how to setup the differential equation needed to map out the total time the Car takes to fall 4000 ft (1219.2m)?

Force due to drag is
Fd=1/2$$\rho$$v2ACd

Where v seems to be the changing variable here. Therefore, force due to drag is a function of velocity in this case?

5. Jan 11, 2010

diazona

In the episode in question, if I remember correctly, they could only get their car up to 120-ish mph, so they adjusted the horizontal distance to the target accordingly... it wasn't much of a physics experiment. But still cool I think the falling car hit first by a second or two.

6. Jan 11, 2010

diazona

Sure, just use $F = ma$, remembering that $a = \ddot{y}$ and $v = \dot{y}$ (dot stands for the time derivative, of course).

$$\frac{1}{2}\rho A C_d \dot{y}^2 - mg = m\ddot{y}$$

oooh, wait - is this a homework assignment?

7. Jan 11, 2010

Staff: Mentor

....and that's if it stays stable and nose-down...

When they meausred the terminal velocity of a falling (not spin stabilized) bullet, it fell sideways and was roughly stable in that configuration.

8. Jan 11, 2010

Staff: Mentor

[googles] 105mph, and with the falling car hitting a full 2 seconds before (308 feet at 105 mph), I wonder what they scaled their distances to?

....and btw, the falling car fell sideways, though oscillating somewhat. Here's the video: http://rippol.com/watch/mythbusters/car-drop-minimyth/?t=3025&v=2169551

9. Jan 11, 2010

danielatha4

They scaled the distance to 2950 feet I believe.

I'm still trying to straighten up that differential equation to figure out an accurate time.

10. Jan 11, 2010

mgb_phys

You can do it easily with a spreadsheet, I get 90% of terminal roughly after 12.5 secs and a distance of 1220m in 25.5 secs
In that time a car doing 140mph would cover about 1450m

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Last edited: Jan 11, 2010
11. Jan 11, 2010

danielatha4

I'll take your word for it. I was gonna try to use the math, not a spreadsheet, but oh well.

if it takes 25.5 seconds to fall then the driving car will pass the x 9.7 seconds before impact. Hmm

And no diazona, this wasn't a homework assignment. Just physics for fun, ha

12. Jan 11, 2010

danielatha4

So the car falling hit the x 2 seconds before the car driving reached it. That means it took 17.2 seconds.

How can you calculate 17.2 seconds from the differential equation?

13. Jan 12, 2010

diazona

As long as you're not trying to leech answers off us ...
That's actually an integrable equation,
$$\frac{1}{2}\rho A C_d \dot{y}^2 - mg = m\frac{\mathrm{d}\dot{y}}{\mathrm{d}t}$$

$$\int_0^t\mathrm{d}t = \frac{1}{g}\int_0^t\frac{\mathrm{d}\dot{y}}{\frac{\rho AC_d}{2mg} \dot{y}^2 - 1}$$
You can do the integral on the right using the substitution $u = \dot{y}\sqrt{(\rho A C_d/2mg)}[/tex], getting $$t = \sqrt{\frac{m}{2\rho gAC_d}}[\ln(1 - u) - \ln(1 + u)]_0^t$$ and if you solve that for [itex]\dot{y}$, taking into account that $\dot{y}(0)=0$, you get
$$\dot{y} = -\sqrt{\frac{2mg}{\rho A C_d}}\tanh\biggl(\sqrt{\frac{\rho gAC_d}{2m}}t\biggr)$$
Then you integrate it again to get
$$y = -\frac{2m}{\rho A C_d}\ln\cosh\biggl(\sqrt{\frac{\rho gAC_d}{2m}}t\biggr)$$
Using mgb_phys's values with y=-4000ft, I get 25.6 seconds, same as the spreadsheet analysis. (No surprise there) Though these calculations are never really exact - there are a lot of things that could have been different in the actual experiment, like a higher mass, lower air density, or a lower drag coefficient. Any of those would reduce the fall time.