2006-10-04 05:39:02 · 5 answers · asked by lizjunocom 1 in Science & Mathematics ➔ Physics
That it crashed at the bottom of the cliff?
2006-10-04 05:47:29 · answer #1 · answered by tallerfella 7 · 0⤊ 0⤋
Height of the cliff and general slope of it down to the point of impact. If there were skid marks at the top of the cliff, indicating an attempt to stop. The distance the car impacted from the base of the cliff, indicating how fast the car was going.
If there were no skid marks and the car was going very fast when it left the top of the cliff, that would indicate possible suicide. (However, a knowledgeable driver or ABS would pump the brakes because rolling friction is greater than sliding friction; so the absence of skid marks would not be conclusive.)
If there were skid marks and the car left the cliff top at a relatively slow speed, that would indicate an accident the driver was trying to avoid. (The presence of skid marks is reasonably indicative the driver was trying to stop...but didn't know the best way to do it.)
Of course, there are other things that would go into that note (like DOA, breathalizer results, deployment of air bags, etc.), but I suspect you are looking for the speed of the car when it drove off the cliff. v(H) = d/t and h = 1/2 g t^2; so t = sqrt(2h/g) and v(H) = d/sqrt(2h/g) = speed off the cliff in meters per sec; where d is the distance of impact from the lip of the cliff, h is the height of the cliff, and g = 9.81m/sec^2 the acceleration of the fall due to gravity.
This was a fun question...puts physics into the context of a real-world scenario.
2006-10-04 06:02:41 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋
the car is no longer on the top of the cliff.
2006-10-04 05:46:20 · answer #3 · answered by seanachie60 4 · 0⤊ 0⤋
That 58m is waaaaay too high for a car to jump safely?
Is that really your question? It's not really much of a question, is it? Seems as if perhaps you forgot to post some additional details?
2006-10-04 05:46:40 · answer #4 · answered by abfabmom1 7 · 0⤊ 0⤋
answer --> 40-one.3 m/s Given H = 40 5 m R = one hundred twenty 5 m g = 9.80 one m/s^2 Equation Vo = R * SQRT { g / [2H] } Vo = (one hundred twenty 5 m) * SQRT { (9.80 one m/s^2) / [ 2 * (40 5 m) ] } Vo = (one hundred twenty 5 m) * SQRT { (9.80 one m/s^2) / [ ninety m ] } Vo = (one hundred twenty 5 m) * SQRT { 0.109 s^2 } Vo = (one hundred twenty 5 m) * (0.330 s) Vo = 40-one.3 m/s
2016-11-26 02:29:57 · answer #5 · answered by Anonymous · 0⤊ 0⤋