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Parabolic motion. air resistance?

  1. Feb 16, 2007 #1
    1. a 150 pound person on a snowboard launches off a 45 degree angle jump at 12 feet per second. what is the peak hight, flight time, horizontal displacment, and vertical displacement.

    3. i tryed to use kinimatic equations to determine all the variables but all were way off[WAY off]. i think this is because i'm not taking into acount friction of snow and air ressistance. [i did the equations about seven times and still came up with the same answers.]
  2. jcsd
  3. Feb 16, 2007 #2
    Welcome to the forums, 403036387. Why don't you show us your work, and we could tell you where you might have made a mistake.
  4. Feb 17, 2007 #3
    Dont think you have to take the air resistance/friction of snow into account. Separate the motion into x and y directions and solve individually. Since the angle given is 45 degrees, your job automatically becomes simpler cause in this case, the v(x) and v(y) are equal.
  5. Feb 22, 2007 #4
    chaoseverlasting, i did just that. i'm still doing something horribly wrong.


    so vx=8.496 and vy=8.496.
    and as far as i know flight time = vy/32.2*2
    so the flight time comes out to:
    which is aproximitley .527.
    then flight distance =flight time*vx, so:
    .527*8.496= 4.483
    if you hit a jump going that fast you would not go four and a half feet. i know. i've hit jumps like that and you fly really far. i re-did the equations with my snowboarder going 40 mph, or 19m/s, [which is what i guess pro boarders might be going if they were booking] and i got that he would fly 36 meters! now thats way too much. i'm really screwing somthing up. maybe i have my formulas wrong?
  6. Feb 22, 2007 #5

    D H

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    You are getting a counterintuitive result because you probably have never gone off a jump like this. The landing is flat. Try tilting the original problem by 45 degrees (horizontal takeoff, 45 degree landing) to get something more akin to what you encounter on the slopes (but 45 degrees is a very steep slope).

    In any case, you did just fine, except cos(45 degrees) is 0.707, not 0.708 (to three digits).
  7. Feb 22, 2007 #6
    why is flight time=Vy/32.2? seems to me that would be apogee.
  8. Feb 22, 2007 #7

    D H

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    Look at his work. He has flight time as Vy/32.2*2, which is correct.
  9. Feb 22, 2007 #8
    Maybe I'm missing something, but the t given by that eqn, would be to the apex, ie when Vy=0 eg Vy(t)=Vy(init)-gt. Then the jumper has the descending portion of the parabola.
  10. Feb 23, 2007 #9

    D H

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    You are missing the factor of two. The equations of motion for this problem are

    [tex]x(t) = v_x t[/tex]
    [tex]y(t) = v_y t - \frac 1 2 a t^2 = (v_y - \frac 1 2 a t)t[/tex]

    The apex occurs when [itex]dy/dt = 0[/itex], or [itex] t = v_y/a[/itex]. The skiier is on the ground when [itex]y(t)=0[/itex], which occurs when [itex]t=0[/itex] (takeoff) and [itex]v_y - 1/2 a t = 0[/itex] or [itex] t = 2 v_y/a[/itex] (landing).
  11. Feb 23, 2007 #10
    Er, beg pardon, I'm not missing anything, just pointing out that only 1/2 the flight time was accounted for. I thought that was sufficient for the OP to work out the problem.
  12. Feb 23, 2007 #11

    D H

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    The OP has had the problem right from the very original post. The full flight time is taken into account (see post #4, t = vy/32.2*2). His problem is that the answer he got, which is correct, is counterintuitive.
  13. Feb 23, 2007 #12

    D H

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    One reason for the counterintuitive results is that the distance is proportional to the square of the skier's velocity. Going a bit faster means going quite a bit longer. This skier was going at a moderate speed:


    Since the distance is proportional to the square of the velocity, one must take care not to go too fast. The next two inspirational posters are very applicable:


  14. Feb 24, 2007 #13
    My bad: :blushing: Im used to seeing the algebra expressed as 2*Vy/32.2, missed the trailing 2.
  15. Feb 28, 2007 #14
    i have no idea what your saying. how do i tilt the equation? and i REALLY don't understand why that would give me a result more similar to what i would encounter on the slopes.
  16. Feb 28, 2007 #15
    What he is suggesting that if the slope falls away from you at an angle
    (45 would be rugged indeed) the distance traveled is gonna be greater because you have a positive initial Y0.

    Looking at that problem:

    Vx(init)=12 ft/sec.

    horizontal displacementat landing=12'/s*t=Sx
    vertical displacement=1/2a*t^2=16*t^2=Sy
    since the angle of the slope is 45, Sx=Sy
    16*t^2=12*t t=0.75 and the horizontal displacement is 9', but the distance down the slope is 9/.707=12.73

    Remember 12'/s is pretty tame, about twice walking speed. Hit it at 20MPH=30'/s and the distance goes up to nearly 40 feet down the slope.
  17. Mar 7, 2007 #16
    Not to sound like a total doof but i'm still a little confused. Those results do sound much more logical. But if i'm launching off of a 45 degree jump that is changing my vy by sending me up, not down, I don't understand why the drop-off equation works so much better. I see that it does, but I don't understand why.
  18. Mar 7, 2007 #17
    Also, I usually have a 45 degree take off, but somewhere around a 35 degree landing. Is there any way to figure that in? Does it even matter?
  19. Mar 7, 2007 #18
    The way the change in the problem was posed was a horizontal takeoff with a slope of 45 degrees falling away from you. Now if you want the slope to be 35 and the jump 45, thats a bit like the two problems rolled in to one. Here we make a ramp like those used in the x games with your spec of 45 degrees, and a face 15 feet high.

    lets take 30 feet/sec to be the takeoff velocity.

    just like before Vy(init)= sin 45* 30:
    and Vx(init)=Vx=21.21 feet per second
    so that the horizontal displacement = 21.21*t

    and the total vertical displacement,

    yo=face of the jump plus the unknown distance down the hill=30+y(?)

    we can relate y(?) to the horizontal distance traveled by tan(35)

    so that tan(35)=y(?)/[21.21*t] ==> y(?)=21.21*t*tan(35)

    then we just use your well known eqn:

    Yf=Yo+Vy(t)+1/2(a)t^2, taking care to keep the signs straight, leads to something like

    since tan 35=.7 this becomes -16t^2+1.7*21.21 *t +15=0 assuming I haven't made any mistakes t=2.62 seconds and ends up 68 feet down the slope. That all make sense?
  20. Mar 22, 2007 #19
    thanks. got it now. 68 feet sounds perfect for a jump like that.
  21. Mar 23, 2007 #20
    you bet. always best when you get a believable physical result with understandable physics in support of. I may of made a small error using 30 instead of 15, but the results should be close in any event.
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