This is just a problem that has been bugging me. This is not me homework.:D It was on a car commercial.

A car company claims they can drop a car 4000 feet and have another car on the ground. I'm pretty sure the car on the ground traveled the same distance. Assuming the car being dropped accelerates at 10 meters per second per second, and the car on the ground is already traveling at a constant speed, can the car on the ground travel faster? How would you work this problem? Is there enough information?

It's just bugging the heck out of me!:confused:

I don't really know what the terminal velocity of a car would be, which is quite internal to the problem :/

How about we assume there is no limit on how fast you can go. Now how would you solve it?

okay so

v^2 = u^2 + 2as

v^2 = 0 + 2*32.15*4000

v = 507 ft/s

that's 154 m/s

Average speed = 154/2 = 77m/s = 0.077km/s = 277km/hour

The car on the ground would have to be going a constant speed of 277km/h or 172 mph

Doable I think!

It would be slower than this due to air resistance. It seems very doable and not really a stunning performance :)

okay so

v^2 = u^2 + 2as

v^2 = 0 + 2*32.15*4000



So what do the variables a and s and u stand for, and why is the equation v^2 = u^2 + 2as?

That's a classic equation of motion for uniform acceleration.

(put "suvat" into google)

a is acceleration, s distance, u initial speed.

So how do you get 32.15 as acceleration?

That's 9.8m/s/s in ft/s/s (a standard result)