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Physics Help

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aoe
Posts: 2947 - Karma: 51
So I'm just doing this for fun, trying to calculate the pontential distance traveled of a jump (I'm think about this for a biking application but im sure it could be used for skiing as well). I havn't done physics since high school so I don't even know where to begin. I would think the following variables would apply: Gravity (-9.8m/s), velocity, mass of object (maybe?), angle of jump, air resistance (friction coefficient for air). I would think you would be able to figure it out from these. anyone know how to do it?
Sep 14 2006 1:29PM
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BurgRider
Posts: 1597 - Karma: 17
thats a pretty complex one. I could do it but i have no motivation at all to do any of that.haha
Sep 14 2006 2:01PM
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BurgRider
Posts: 1597 - Karma: 17
its basically the same as the problems where they ask how high the ball goes if thrown but you just factor in that it is not going vertical and that it has already been accelerated to its maximum velocity in the begging of the equation
Sep 14 2006 2:03PM
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aoe
Posts: 2947 - Karma: 51
ok cool. what if you are going off a jump with a gradual lip (i.e. constantly increasing angle)? Do you just go by the angle at which it leaves at or is there a further equation to factor in the curve?
Sep 14 2006 2:15PM
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flippintricker
Posts: 2408 - Karma: 9
the angle at which it leaves. also, for skiing, youd have to remember that the landing will be lower than the lip
Sep 14 2006 2:39PM
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aoe
Posts: 2947 - Karma: 51
well there are a lot of variables but the landing might not be lower then the lip in relative terms. for example say you are comming down a 30 degree slope and hit a jump with a 30 deg lip. you are essentially dropping from a flat surface in absolute terms. so as long as you are landing on that same 30 deg slope then it would be the same as travelling from flat hitting a 30deg kicker and landing flat. So is it definate that mass of the objet and change in angle of the jump have no effect?
Sep 14 2006 2:46PM
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aoe
Posts: 2947 - Karma: 51
I think this is the right formula:

H =Vy T - 0.5g T^2
Sep 14 2006 2:47PM
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aoe
Posts: 2947 - Karma: 51
then Plug Time into x = Vx T
Sep 14 2006 2:56PM
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BurgRider
Posts: 1597 - Karma: 17
not necessarily
Sep 14 2006 3:13PM
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BurgRider
Posts: 1597 - Karma: 17
Im not sure about that one...im going to chicago right now but. im determined to figurte this out when i get home
Sep 14 2006 3:16PM
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SkiFakie614
Posts: 541 - Karma: 10
This my friends, requires a differential equation.

You have the resistance of the air which depends on the speed of the bike/skier... Air resistance is not like friction where you only need a normal force and coefficient. Air resistance relies on frontal area and velocity; which changes as you travel through the air continuously. You also have the inrun and the amount of kinetic energy burned off to friction between your wheels and the ground.
Sep 14 2006 5:47PM
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SkiFakie614
Posts: 541 - Karma: 10
So now the question is... who besides me has taken university physics and a differential calculus class?
Sep 14 2006 5:48PM
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ECfreeski3
Posts: 12019 - Karma: 126
thats mad easy to calculate tho
Sep 14 2006 5:50PM
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mauii
Posts: 6166 - Karma: 32
if your going to take air resistance into account, then there's the fact that a peron.. on skis or a bike.. is probably moving while in the air.. ie: constantly changing face surface area. Then if the wind is blowing... yeah.. you might just wanna neglect air resistance for this one.
Sep 14 2006 5:58PM
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RKF
Posts: 1527 - Karma: 342
yeah i was about to suggest the same thing. or possibly find the maximum air resistance and just treat it as a constant because i figure its better to go a little bigger than hit the knuckle.
Sep 14 2006 6:01PM
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aoe
Posts: 2947 - Karma: 51
ok so we are offically neglecting air resistance on this one. I think I got it to work. Basically I came up with a skiier/biker leaving a 20deg lip at 50kmh will travel 50ft anyone wanna check that?
Sep 14 2006 8:45PM
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cobra_commander
Posts: 16909 - Karma: 4,686
well I will just say that a lip at 45 degreess will travel further, and theoreticly a 90 degree lip will luanch you just as far as a 10 degree lip.
Sep 14 2006 9:03PM
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SkiFakie614
Posts: 541 - Karma: 10
50kmh going 50ft in the air?

... units?!
Sep 15 2006 3:11AM
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aoe
Posts: 2947 - Karma: 51
I already did the conversions. I like working in feet for distance
Sep 15 2006 6:47AM
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Intergalactic
Posts: 1858 - Karma: 47
gravity is actually -9/8m/s^2
the seconds are squared
Sep 15 2006 8:51AM
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flippintricker
Posts: 2408 - Karma: 9
im in astrophys, but just starting second year, so no course devoted entirely to differential equations yet
Sep 15 2006 11:27AM
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aoe
Posts: 2947 - Karma: 51
I've taken differential and integral calc. no physics though
Sep 15 2006 12:04PM
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Itsbackfliptime
Posts: 18901 - Karma: 75
ME ME ME! I sucked ass at physics though. C-. I got a B+ in Diff Eq though.
Sep 15 2006 1:00PM
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treesmann
Posts: 2675 - Karma: 49
i got a 4 on the ap physics last year. mass is not needed but im sure i could figure that out.
Sep 15 2006 1:40PM
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thecoyote
Posts: 1937 - Karma: 14
do you mean an 80 degree lip will launch you just as far as a 10 degree lip? a 90 degree lip will send you up, and right back down like a quaterpipe.
Sep 15 2006 2:17PM
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^nate^
Posts: 1263 - Karma: 560
a formula could be derived using dimensional analysis. take in all the factors, like you listed and put their units to powers of symbols and solve for the exponents creating an equation
Oct 13 2007 3:03PM
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jacqui-la-fripui
Posts: 1119 - Karma: 12
exactly i was gonna say fuck air resistance
Oct 13 2007 3:50PM
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t-roy
Posts: 4041 - Karma: 52
this is an interesting thing to do for fun. i think you would be best off jsut treating it as a simple trajectory problem. air resistance and weight aren't going to be huge factors and would make it much more difficult. but if you want it really challenging, don't forget to consider wind, bagginess of clothing, width of skis, air density at different altitudes and relative humidity.
Oct 13 2007 4:28PM
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TheQuailman
Posts: 7382 - Karma: 147
Start with the horizontal distance you need to clear. Call it X.

The time the car will be in the air is
T = X / (V cos A)
A is the ramp angle from horizontal and V is the initial velocity.

One must also satisfy
V sin A = g (T/2)
because that it the vertical velocity component change in time (T/2)
Therefore
T = 2 V sin A/ g

Putting the two equaqtions together,
X / (V cos A) = 2 V sin A / g

X = 2 (V^2/g) sin A cos A = (V^2/g) sin 2A

You can use that equation to find combinations of V and A that work.

Note that sin 2A cannot exceed 1, and it has its greatest value when
A = 45 degrees. That will give you the lowest value of V needed for a given gap X.

The same equation comes into play when you are talking about the range and optimum launch angle of a thrown projectile.

A long jumper will also travel farthest if his/her takeoff angle is about 45 degrees.

(I found this on NS, don't know who posted it - hope it helps)
Oct 13 2007 6:07PM
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jmason
Posts: 1595 - Karma: 136
I'll help out in a while, I've just been working on this! I intend on putting it to use for both biking and skiing asap!
Oct 13 2007 6:41PM
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dgcarbs
Posts: 438 - Karma: 11
because gravity is measured by acceleration not velocity
Oct 13 2007 6:41PM
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bigdaddytmac
Posts: 1448 - Karma: 14
solving a differential equation is not necessary, the velocity changes a small enough amount that you could neglect that. Air friction could be neglected all together and it would be a simple calculation.
Oct 13 2007 9:05PM
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jmason
Posts: 1595 - Karma: 136
I have

Air resistance can be neglected. I was just looking through my old stuff, and I have a 2 page note on exactly how to do this, because I worked on it so I could put it into use when building. I'll upload it eventually for you. It's really well laid out and easy to follow, and all you'll have to do is plug-n-chug
Oct 13 2007 11:11PM
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sk8almost13x
Posts: 498 - Karma: 10
2 words mad confusing
Oct 14 2007 12:00AM
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comeatmebro
Posts: 10720 - Karma: 66
def. dont have the time for this right now. but 2 things, mass does not matter, and get rid of air resistance wayyy to complicated unless you are willing to spend a lot of time with changing values etc. idk i didn't read the responses but hope that helps a little....
Oct 14 2007 12:06AM
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jacqui-la-fripui
Posts: 1119 - Karma: 12
pics or it didnt happen
Oct 14 2007 9:08PM
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AMG.
Posts: 867 - Karma: 21
the weird part is that i'm in grade 11 and had allready thought of all those variables. however, i dont know how to make the equation, but then again, ive never taked a physics or calculus class in my life.
Oct 14 2007 9:16PM
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