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wsclimber
Dec 15, 2004, 12:10 AM
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Found a new crag and threw a rock off the top, it took 5 seconds for the rock to hit the ground. Can someone help me with the math to figure out how tall the crag is? Thanks in advance.
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sandbag
Dec 15, 2004, 12:14 AM
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In reply to: Found a new crag and threw a rock off the top, it took 5 seconds for the rock to hit the ground. Can someone help me with the math to figure out how tall the crag is? Thanks in advance.
So that was you tossing rocks off of the top, man, you better watch your A$$ throwing stuff off like that...........
you figure it out: 9.81m/s/s and thats wher eyou start tough guy.
go troll elswhere
anyone knows trig would work better.
:P
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cactusjack
Dec 15, 2004, 12:17 AM
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ACCEL=9.8M/S
VELOCITY=9.8*TIME+C1, C1=0 IF YOU DROPPED AND DID NOT THROW THE ROCK
POSITION=9.8*T^2/2+C1*T+C2
POSITION=9.8*T^2/2, C1 AND C2=0
POSITION=122METERS=402FT
MMMM, I WONDER IF THAT RIGHT
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sandbag
Dec 15, 2004, 12:19 AM
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doesnt matter if you drop it or throw it bro, 9.81 is the magic number.
given most people throw upwards a little when trying to get distance in the horizontal vector, youll get a lil variance, problem is is he talking 5 second to visual impact or till the sound, aha!
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cactusjack
Dec 15, 2004, 12:22 AM
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OH, YEAH YOU CAN ALSO SUB IN 32 FT/S IN PLACE OF 9.8 M/S DEPENDING ON IF YOUR METRIC OR STD.
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tchamber
Dec 15, 2004, 12:22 AM
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9.8t^2/2 = ht in this case
122.5... meters of course
~366 feet
edited to say-- same result as cactus jack... some variables as far as surface area and wind resistance aren't accounted for, but the more likely source of error would be the accuracy of timing...
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cactusjack
Dec 15, 2004, 12:24 AM
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AHH good point Sandbag about the visual vs audio. However if he throws it, then c1 should be set to the initial velocity at which he threw it, changing the finnal equation. adding a +c1*t component.
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sustainedclimber
Dec 15, 2004, 12:26 AM
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Assuming you just let go of the rock and it had no initial velocity, and also assuming you were in a vaccum, then you want to use the equation, Height=acceleration of (1/2) * gravity * time^2. gravity= 9.8 meters/(seconds)^2 and your given time of 5 seconds...the answer would be 122.5 meters or 401.8 feet. However, we must take into consideration the amount of time it takes the sound of the rock hitting to reach you at the top of the cliff. The velocity of sound in air is ~340 meters/second at sea level, so the number is actually less than that 400 foot estimate. I don't really feel like doing the math out right now. If I had to guess, I'd say it's about a 350 foot cliff from the numbers you are giving me.
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wsclimber
Dec 15, 2004, 12:27 AM
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Thanks for the help! Guess I'll need to tie two ropes together when rapping (even at 4 secs its 256 ft tall).
And there isn't even a trail near by, never mind someone at the base (where we just had been, and could see) so please don't kick my ass for dropping a rock....
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sandbag
Dec 15, 2004, 12:28 AM
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wrong.
the instant a projectile leaves the safety of its anti gravity position, its being acted upon with the acceleration due to gravity, so the variance is slight, and if its airfoil shaped i may actually rise, but its pretty fair to say unless you really arc the up vector when throwing its neglible for this problem....you could be anal and add say 2 meters to position 0 for the height of the person and the arc of the throw
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noshoesnoshirt
Dec 15, 2004, 12:33 AM
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In reply to: doesnt matter if you drop it or throw it bro, 9.81 is the magic number. given most people throw upwards a little when trying to get distance in the horizontal vector, youll get a lil variance, problem is is he talking 5 second to visual impact or till the sound, aha!
No no, it does matter.
X = Xo + Vo*t + 1/2*g*t^2
where v = initial velocity. If initial velocity is not zero the time necessary to travel the distance X will not equal the time necessary to travel the same distance X with an initial velocity of zero. Just do the math;
let Xo=0, Vo=0, X = 1/2*g*t^2;
now let Xo=0, Vo=10, X =10*t + 1/2*g*t^2.
Not a huge difference considering the measuring technique, but a difference nonetheless.
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sandbag
Dec 15, 2004, 12:49 AM
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screw the vertical velocity and just go use trig to figure it out, more accurate, almost no guessing then
just take a protractor and find a 90(plumb bob will do it) the opposite angle, known distance from the base and calculate the height.
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noshoesnoshirt
Dec 15, 2004, 12:59 AM
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screw the math, rap it.
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sandbag
Dec 15, 2004, 1:04 AM
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In reply to: screw the math, rap it.
take a known length of static line, rap it then report back on if you had to use it more than once or twice even......... :roll:
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timd
Dec 15, 2004, 6:45 AM
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Your giving me a headache, just climb the f----n thing and figure it out by rope length :lol:
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coldclimb
Dec 15, 2004, 7:07 AM
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Kids, this is what happens when you go to college! :shock: :shock:
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guangzhou
Dec 15, 2004, 7:13 AM
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Was this an actual timed 5 seconds, or was it one of those 5 Mississipi time count? :shock:
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curt
Dec 15, 2004, 7:32 AM
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If you throw the rock or drop the rock it makes no difference in how much time it will take the rock to reach the ground. Assuming, of course, that the thrown rock is thrown parallel to the ground surface and there is no aerodynamic lift component involved.
Curt
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scythide
Dec 15, 2004, 7:42 AM
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Also, did you watch the rock hit the ground or did you hear it? Cause if you relied on sound, you'd have to factor back in the time it takes for the sound to travel to your ears.
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boulderinemt
Dec 15, 2004, 2:39 PM
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In reply to: Kids, this is what happens when you go to college! :shock: :shock: |
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bbecker
Dec 15, 2004, 3:24 PM
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I didn't learn that stuff in college!
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sticky_fingers
Dec 15, 2004, 3:24 PM
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Why don't you just tell us where this new (secret?) crag is and then maybe somebody will tell you the specs about it (height, number of routes, difficulty of routes, etc). May not be that secret, or maybe somebody else can go wth you to measure. Don't forget your Bosch! :P
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sustainedclimber
Dec 15, 2004, 4:27 PM
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In reply to: If you throw the rock or drop the rock it makes no difference in how much time it will take the rock to reach the ground. Assuming, of course, that the thrown rock is thrown parallel to the ground surface and there is no aerodynamic lift component involved.
Yes!! I love assumptions!
-Josh
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cactusjack
Dec 15, 2004, 5:07 PM
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Ok I got bored at work:
Disclaimer, basic understanding of algebra, dynamics, physics or calculus is required.
Assumptions: rock was dropped not thrown, yes it makes a difference!, of exactly Vo*t, how fast you threw it downward times the length of time: i.e. 10m/s down, it fall for 5 secs, adds 50m. Air drag was not taken into effect, since the rock most likely did not fall long enough to reach terminal velocity, and since most rock will tilt in mid fall to reduce their air drag. But for those of you who want to know: http://www.grc.nasa.gov/WWW/K-12/airplane/termv.html you factor it in. Time measured was from when dropped to when you herd it.
So here goes:
G= gravity 9.8m/s
Speed of sound is 348m/s
H (wall height in meters)=
T (Total time of Fall )=5 sec
Tf (time for fall)
Ts(time for sound to travel back)
T=Tf+Ts
H=(G*Tf^2)/2 , FROM ROCK FALLING
H = 348*Ts , FROM SOUND TRAVELING UP
The genric equation to solve for Tf is:
(G/2)*Tf+348*Tf-348*T=0
solving using, T=5, Tf=4.7sec
Solving H=(G*Tf^2)/2, using Tf=4.7sec
Answer H=108m ~355ft
Sustainedclimber’s climbers estimate was dead on.
Please do not bash this post.
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cactusjack
Dec 15, 2004, 5:11 PM
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correction:
(G/2)*Tf^2+348*Tf-348*T=0
in case you have future crag hts you want to figure out.
Last peep out of me, on this topic.
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