3 ms·
Let's see if I remember my high school physics. Making all sorts of assumptions and simplifications here: F = mv / dt (equation of impulse for zero final veloc
by a-priori 13y ago
Let's see if I remember my high school physics. Making all sorts of assumptions and simplifications here:
F = mv / dt (equation of impulse for zero final velocity)
F = mg (newton's laws)
mg = mv/dt
g = v/dt
9.8 = v/0.5
v = 4.9
Therefore, the car can be travelling no more than 4.9m/s at impact to survive.
mgh = 0.5mv^2 (initial gravitational potential energy = kinetic energy at impact)
gh = 0.5v^2
h = (0.5v^2) / g
h = (0.5 * 4.9 * 4.9) / 9.8
h = 1.225
So, assuming that it takes 0.5 seconds (a wild ass guess) for the car to go from freefall to rest, the car can withstand a drop of 1.2 metres without deforming the roof.
Mind you,
- mikeash 13y agoOuch, units, please! Leaving the units out of your physics math is like, well, not wearing a seat belt when driving a car. Anyway, I'm afraid that 0.5 seconds to come to rest is way too long. Toss it into the equations of motion and: d = 1/2 at^2 d = 1/2 9.8m/s^2 (0.5s)^2 d = 4.9m In other words, decelerating at 4 gees for half a second means you decelerate over nearly five meters. I think we're better off starting at the other end of the stick and making a wild guess that the roof can move 10cm without being permanently deformed. Then: d = 1/2 at^2 0.1m = 1/2 4 * 9.8m/s^2 t^2 t = 0.0714s To figure out the height, we can notice that a and t will be inversely proportional for any given velocity (double the time, halve the acceleration needed), and so, for a given initial or final velocity, the subexpression at^2 is directly proportional to the change in acceleration. In other words, if we assume that the roof can withstand 10cm of deflection, you can drop the car from 40cm up.
- a-priori 13y agoThanks, much appreciated. Sorry about the lack of units. If it's any consolation, I included them in the text surrounding the calculations. :)