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04212011 12:47 AM 7 years ago  Post 1 
GyroFreak rrProfessor Orlando Florida ...28N 81W  My high school and college physics memory is shot. Here is the question.
 

04212011 01:14 AM 7 years ago  Post 2 
ssmith512 rrKey Veteran Indianapolis, IN USA  Oh dear. Equations of motions. I think it is v^2 = u^2 + 2asv=final velocityu=intitial velocity a=acceleration (g's) s=distancesolving for "s" I get 140.08 feetIt's been a long time so I may be wrong Steve  

04212011 02:04 AM 7 years ago  Post 3 
rcjon rrVeteran Macon, GA  That's what I got by:
average velocity = (vi+vf)/2distance = (average velocity) (time)  

04212011 06:57 AM 7 years ago  Post 4 
GyroFreak rrProfessor Orlando Florida ...28N 81W  Thanks guys, I think I need to brush up on my old physics books. It's a question my grandson asked me. "If person is falling at terminal velocity and you could apply a constant 3 G force, how long would it take to stop safely in some type of crushable foam."
 

04212011 05:50 PM 7 years ago  Post 5 
fla heli boy rrKey Veteran cape coral, florida  I am VERY rusty with physics, but it seems to me there has to be a place for MASS in that equation.....  

04212011 05:55 PM 7 years ago  Post 6 
TrexRookie rrKey Veteran San Francisco, CA  I'm not sure mass has anything to do with it... The question asks for distance. If this were a question of force then sure, but it doesn't look like this is the case. I may be wrong tho, so I'm not totally sure about that.. haha+1 for obscure physics advice.  

04212011 06:09 PM 7 years ago  Post 7 
smh20502 rrNovice Houston, Texas  Kind of a trick question since the g rating of a safe landing is very high (better than 30g's). Where is this variable?Common sense isn't so common. Yes, there are stupid questions  

04212011 06:48 PM 7 years ago  Post 8 
rcjon rrVeteran Macon, GA  As Mr. Newton showed us, if acceleration is constant, mass does not matter.
 

04212011 07:08 PM 7 years ago  Post 9 
fla heli boy rrKey Veteran cape coral, florida  OK  think about this. A bowling ball dropped next to a ball of the same size made of lightweight styrofoam. Surely mass HAS to matter if the same amount of force is being applied to stop/decelerate the objects.
 

04212011 07:56 PM 7 years ago  Post 10 
MartyH rrProfessor USA  I think mass is covered by the 3G's.  

04212011 07:57 PM 7 years ago  Post 11 
rcjon rrVeteran Macon, GA  Correct that the foam is not going to stop a light object in the same distance as a heavy object.
 

04212011 08:06 PM 7 years ago  Post 12 
bbaxter rrApprentice Central Illinois  If you apply the same force to two objects, the lighter one will accelerate faster. However, the original question postulated the acceleration (deceleration) that was applied.So, if you apply a 3G deceleration (negative acceleration) to both a truck and a car, if they are traveling at the same velocity, they'll take the same distance to stop. However, a greater force will be applied to the truck since force is proportional to mass and acceleration. F = MA.Not counting any influence from the air, a bowling ball and a styrofoam ball will fall together because they are both in a constantacceleration field. Gravity really isn't a force, it's an acceleration. Weight is the what we call the force we experience.We use scales that measure force and have the readout calibrated to read mass, so they basically "take out" the acceleration of gravity.If you use a scale and see that you weigh 100 Kg, you actually experience a force of 980 Newtons (100Kg x 9.8 m/s/s).  

04212011 08:43 PM 7 years ago  Post 13 
AirWolfRC rrProfessor 42½ N, 83½ W  Mass doesn't mattere here.
 

04212011 11:59 PM 7 years ago  Post 14 
GyroFreak rrProfessor Orlando Florida ...28N 81W  Thanks AirWolfRC, that is a very clear explanation of my stated question. Several others got the same answer, so thanks guys.
 

04242011 05:38 PM 7 years ago  Post 15 
AirWolfRC rrProfessor 42½ N, 83½ W  There was a leap to the final formula I left previously so I'll explained it a bit more;Mass doesn't mattere here.Only velocity since you state that the deceleration is 3 G's. 3G's is 3 x 32ft/sec² = 96 ft/sec². Time to stop is 164 / 96 = 1.71 seconds.So, to be redundant, you decelerate at 96 ft² for 1.71 seconds That's A x T which gives final (start) speed.Average speed = 1/2 of that Average speed = 1/2 x A x TDistance = average speed x time Distance = (1/2 x A x T) x T Distance = 1/2 x A x T² 1/2 x 96 x 1.71² = 140 ft  

04242011 06:16 PM 7 years ago  Post 16 
Justin Stuart (RIP) rrMaster Plano, Texas  To do an approximation in your head:Assume g is 10m/s^2. Initial velocity is 50m/sForce is 3g or 30m/s^2. What this means it that the object will slow down at a rate of 30 meters per second every second.After the first second, the object has decreased in velocity by 30m/s.After slightly more than 2/3 of the 2nd second, the object has stopped.So the time is approximately 1.7 seconds.Distance is given by the average velocity multiplied by time. Initial velocity is 50m/s. Final velocity is 0m/s. So average velocity is 25m/s25m/s x 1.7 seconds is equal to 25 + 12.5 + 2.5 + 2.5 = 42.5 meters1 meter is equal to approximately 3.3 feet.42.5 x 3.3 = 42.5 + 42.5 + 42.5 + 4.25+ 4.25 + 4.25 = 40 x 3 + 2.5 x 3 + 12.75 = ~140 feet.Not an exact answer, but easier to do in your head than using a second degree polynomial equation: Vi^2  Vi^2 +2at = 0Avant RC  

04242011 06:32 PM 7 years ago  Post 17 
AirWolfRC rrProfessor 42½ N, 83½ W  Do this in your head ? ! ? ! ?25m/s x 1.7 seconds is equal to 25 + 12.5 + 2.5 + 2.5 = 42.5 meters1 meter is equal to approximately 3.3 feet.42.5 x 3.3 = 42.5 + 42.5 + 42.5 + 4.25+ 4.25 + 4.25 = 40 x 3 + 2.5 x 3 + 12.75 = ~140 feet.  

04242011 06:38 PM 7 years ago  Post 18 
Justin Stuart (RIP) rrMaster Plano, Texas  Do you have another way to multiply 42.5 x 3.3 in your head?I look at it as 40 x 3 = 120plus 2.5 x 3 = 7.5plus 4.25 x 3 = 12.75 = ~140Maybe there is a better way to do this (other than to break out a calculator which is sortof cheating)?Avant RC  

04292011 05:11 AM 7 years ago  Post 19 
GyroFreak rrProfessor Orlando Florida ...28N 81W  In my younger days many many years ago I was a slide rule cripple, meaning I couldn't do (or wouldn't do) any math in my head, even simple multiplication. (Course you still had to be able to add and subtract without the slide rule.) Now I am a handheld calculator cripple for the same reason.I think about the hereafter. I go somewhere to get something, then wonder what I'm here after ?  

04292011 05:52 AM 7 years ago  Post 20 
Justin Stuart (RIP) rrMaster Plano, Texas  Throw down your crutch and walk!!Avant RC  

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