# Physics/math in car commercial

You have probably seen the commercial for a car (Nissan?) where they drop a car 4000 feet from a helicopter at the same time a car on the ground is racing toward the target.

My questions:
[ol]
[li]How long would it take a car falling from 4000 feet to reach the ground?[/li][li]What speed (in MPH) would a car on the ground have to travel to cover the same distance in the same amount of time (Assuming, as shown on the ad, that the car is allowed to get up to speed before crossing the start line)?[/li][/ol]

(I don’t know how to insert the Greek letters used in physics in these posts, so ‘de’ means ‘delta’, as in ‘a [small] change’)
(de)y=change in height=4000 feet in meters
v[sub]0y[/sub]=initial vertical velocity
a=acceleration (due to gravity in this case, so ~9.8 m/s[sup]2[/sup])
t=time
(de)y=v[sub]0y[/sub]t+.5at[sup]2[/sup]
(de)y=0t+.5at[sup]2[/sup]
(de)y=.5at[sup]2[/sup]
(2(de)y/a)[sup].5[/sup]=t
(21219 m/9.8ms[sup]-2[/sup])[sup].5[/sup]=t
t=~15.8 seconds

As for the second part of your question, I’m not quite sure what you’re asking. If the helicopter is moving when it drops the car, then the car on the ground will have to match the helicopter’s initial horizontal velocity to keep pace with the dropped car. If you’re asking about the ground car covering the same distance that the air car fell, that’s simply 4000 feet/the ground car’s velocity.

If you disregard air resistance, I get about 15.5 seconds before the car hits the ground. Car hits ground at 344 mph.

To cover 4000 ft in 15.5 seconds, the other car would have to be travelling at 175mph.

Maybe someone can refine those numbers with air resistance taken into account.

Thanks. The way they show it on the ad, the helicopter is hovering over a taget at 4000 feet, while the car is travelling on a road toward the same target. When the car crosses the line marking the 4000 ft. distance, the car is dropped, and we see the ground car crossing the target a split second before the falling car hits.

So, given your answer, what does 4000 ft/15.8 secs translate to in MPH.

Thanks,
Math impaired FBG

Ignoring friction:

Potential energy = mgh
kinetic energy = .5mv^2 (v = velocity when it hits ground, PE = KE)

v = at (a = gravity, which is constant)

solve for t:

t = sqrt(2h/g) = sqrt(2x1220/9.8) = 15.8 sec

to find the speed of a car on the ground, just use d = vt:

v = d/t = 253 f/s = 173 mph (assuming helicopter is stationary)

It would be far more impressive if both cars started from a standstill (which is what I thought was being implied until I watched the commercial more carefully).

This would require a car that could accelerate at greater than g (9.8 m/s[sup]2[/sup]) for the whole 4,000 foot distance (ignoring air resistance). At this acceleration, the car would be traveling in excess of 346 mph.

So I guess that’s not realistic.

I wonder if a car that can accelerate at greater than g is even possible. I don’t believe that it is. You’d need a coefficient of friction between the tires and the road of greater than one. One of those rocket or jet engine cars should be able to do it, though.

I think the question is “can this be for real”? And to answer that we have to find the terminal velocity of the dropped car. It all the equations presented air resistence is ignored, but in real life it is a huge factor. The terminal velocity of a human body upright and spread eagle horizontal is significant. I can’t give numbers, perhaps a sky diving Doper can help me out there. I don’t where to begin to find out the terminal velocity of a falling car at any angle. My point is, given maximum air resistance, and a good running start for the car on the ground, it may be possible for the car on the ground to out-run the falling car. The ad was very misleading all the same. I would love to give the head of the marketing department an opportunity to try this, and if the CEO who approved this was in the falling car that would be a plus.

Here’s a Java applet from the NASA website that calculates terminal velocity given input weight, area, and drag coefficient. I don’t really know what any of these are, but I’m guessing that the drag coefficient is around 1, the cross-sectional area is around 100 square feet, and the weight is around 3500 pounds. That gives a terminal velocity of about 170 ft/sec, or only about 115 mph.

I’ve got a spreadsheet around somewhere that’ll calculate total time to drop accounting for air resistance (which is what the OP asks), but I don’t really know the weight and area of the car in question.

For shorter distances than 4000 feet it is certainly possible. A top fuel dragster covers a quarter mile from a standing start in about 4.5 seconds, and emerges doing well over 300 mph. That’s an average acceleration of about 3 g. Actually, it doesn’t accelerate constantly over the quarter, of course. The wiki article on top fuel suggests that they pull about 5 g in the early part of the run, accelerating from 0 to 100 in about 0.8 seconds.

My hat is off to the folks who took the time to work up the technical answers to FatBaldGuy’s questions. It was a flashy stunt, with the illusion of perfect timing and great danger. :dubious: I salute the computer editors and other technical folks, but of course it wasn’t real. Heck, I don’t even believe that a car dropped from 4000 ft. would land perfectly wheels-down.

If it had happened the way it appears to happen, we’d be asking things like, “How many drop cars and stunt drivers did they sacrifice to get it just right?” :smack:

I think right as the vehicle hits the ground they flashed a disclaimer at the bottom of the screen. Too hard to read because of the way the image is setup and too quick in anycase. I didn’t do the calculations but assumed off the top of my head that it was possible with terminal velocity factored in but still way too risky to try (cross breezes, helicopter not being exactly above the target, aerodynamics of the car causing it to shift horizontally, etc). I’ve been trying to catch the disclaimer but still can’t catch it. I have no doubt it’s all video wizardry, but I’d love to see what the disclaimer says. It’s much longer than something like, “Not real, do not attempt” or “Stunt driver used, do not attempt.”

The car will land wheels-down if the center of gravity is below the center of pressure with the car falling wheels down, and if the center of gravity of the car is directly under the center of pressure in the horizontal plane.

Almost no cars meet that criterion. Most are nose heavy, because that’s where the engine is. That would tend to make them fall nose first. But they’re also not perfecly symmetrical aerodynamically. However, I’ve seen some cars dropped from airplanes, and they seemt to be reasonably stable when they fall. Often they dip back and forth like a leaf falling. So I think it’s possible they could have dropped it and have it land pretty much on its wheels.

“Vehicle shown with optional equipment. Rating achieved using premium fuel”

At the beginning of the commercial, it says
“Based on horizontal drop. Aerial sequence simulated.”

then
“Professional driver on closed course. Do not attempt.”

then the message above at the point of impact.

I contribute nothing except to note that:

1. The stunt is accomplished at the Inyokern Airport, a small airport located in the Indian Wells Valley, about 120 miles north of LA, at the East Base of the Sierra Nevada (the peaks of which you see in the background). It exists primarily to serve the Naval Weapons Air Station - China Lake, and the neighboring support community, Ridgecrest. The tall, pointed peak in the center of the background is Owens Peak, which stands about 8000 feet ASL.

2. I grew up in Ridgecrest, and know that scenery very well. Lots of car commercials get filmed there. This was one of the most innovative.

3. There is almost always a wind blowing down off the Sierras. There would be no way to do the trick for real.

Also by using the correct lens on the camera, it can make the two cars appear to be just this far apart, when in fact they are 100 feet or more apart.

Wait, I need this established:

Was, or was not, a Lexus actually dropped from a helicopter and smashed to smithereens in the making of this commercial?

I think the important thing is that it is plausible. That the time it would take for the car to run that distance at top speed is less than what it would take it to free fall that same distance.

I very much doubt that it was actually done as shown for all the reasons already explained.

Did they drop a car from that height to measure the free fall time? Why not? The mythbusters have done it a couple times and their budget is certainly less than that of Toyota.

Did Toyota actually drop a car from the helicopter? I’m not sure, but I can tell you that cars/trucks + aircraft seem to have some odd fancination for ad guys.
MG dropped an MGB on a pallet with a parachute for a TV ad. There was a great picture in Car and Driver of the take where the chute did not open. It was not a very tall car.
Ford did a Ranger truck ad where they pushed a Ranger out of an airplane, with a group of skydivers. Somewhere at home I have a video tape of the first two attempts. The chutes did not deploy and both trucks went kersplat.* Ever seen a 1 foot tall pickup?
Recently there was a Land Rover ad where they loaded a Land Rover into an airplane and used the Rover’s navigation system to navigate the airplane.

*FTR and it might make a difference in your calculations, the Ford Ranger had a tiny drouge chute open. This stabilized the truck in a slight nose down attitude. It fell at the same speed the skydivers did when they were in the typical arms and legs spread position. Isn’t that about 125 MPH?

I noticed when I saw this commercial during the football games yesterday (the single commercial that wasn’t the rock-em-sock-em robots commercial) that there was fine print at the bottom of the ad saying something to the effect that the “falling car is simulated”. So, no, they did not.