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Projectile lands on an incline.

  1. Jan 12, 2014 #1
    1. The problem statement, all variables and given/known data

    My problem looks like this:
    http://i.imgur.com/3M9wE43.png

    I need to find the point of impact and the time the body traveled.

    2. Relevant equations

    3. The attempt at a solution

    I tried to find the time for 25 meters and with that the vertical velocity and height.
    Ended up getting: t at 25m = 1.96 s , h=16.83 m vi=-1.03 m/s
    I tried this: hf=vx*sin37*t = 7.67 m/s*t , it should mean that for every seconds the slope rises 7.66m in height.
    After that I set up: hf = h + vi*t - 1/2*g*t^2 and got:
    7.67*t = 16.83 - 1.03*t - 4.905*t^2
    -4.905*t^2 - 8.7*t + 16.83 = 0
    4.905*t^2 + 8.7*t - 16.83 = 0

    The positive t was 1.17 s

    Total time = 1.96 s + 1.17 s = 3.13 s

    Height:
    h = vi*t - 1/2*g*t^2
    h = 18.20 m/s * 3.13 s - 1/2 * 9.81 m/s^2 * (3.13s)^2
    h = 8.91 m

    distance = vx*t
    distance = 12.75 m/s * 3.13 s = 39.9 m

    My problem is that the height and distance doesn't match, the tangent of the angle is: height/distance = 8.91/39.9-25=8.91/14.9=0.5979 , with arctg I get around 31 degrees, shouldn't I get 37 degrees here? This is what bothers me. Is the method I used correct? If not, what is the method I should use to solve it?
     
    Last edited: Jan 12, 2014
  2. jcsd
  3. Jan 12, 2014 #2

    SammyS

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    Why use 3.13 sec. for the time to find the height?

     
  4. Jan 12, 2014 #3
    Shouldn't the path of of the trajectory be the same as when I'm solving it without slope?
    Wouldn't solving the height throught tg(37)*14.9 change the path?

    Clear me up if I'm wrong, cause that is the part that bothers me the most about this problem.
     
  5. Jan 12, 2014 #4

    SammyS

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    Yes, of course you're right about that.

    You could use the equation for hf from the ramp, along with t = 1.17 s.


    But looking at you're initial work, how did you get vx and vy ?
     
  6. Jan 12, 2014 #5
    Try the following. Assume the projectile lands on the ramp a distance d horizontally from the base(where the ramp starts) and a distance h higher than the base. d and h are related by the angle 37 degrees. Now do the standard projectile equations d(vertically) = v(0)t = (1/2)at^2 and d(horizontally) = vt. If I say much more I would work the problem.
     
  7. Jan 12, 2014 #6
    Vy=sin(55)*v and Vx=cos(55)*v v is 22.22 m/s

    What to do with the 25 meters? I know how to solve it when the projectile is fried from the bottom of the ramp.

    tg(alpha)=Sy/Sx
    tg(alpha)=(vy*t-gt²/2)/(vx*t) and so on, but where do the 25m fit into all that?
     
  8. Jan 12, 2014 #7

    SammyS

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    moria,

    It looks like you should use tan(37°), not sin(37°) for the ramp, hf.

    Otherwise, your method appears to give a good result.
     
    Last edited: Jan 12, 2014
  9. Jan 12, 2014 #8
    From the initial velocity of the projectile, you calculate the vertical, Vv, and horizontal velocity,Vh, knowing the angle of course.
    The horizontal equation would be 25 + d = Vh * t
    Now write the vertical equation. Remember that d and h are related by the angle of 37 deg.
     
  10. Jan 12, 2014 #9
    With the risk that I might write bullsh*t:
    25+d=vx*t
    d=vx*t-25

    tg(37)=h/vx*t-25
    h=tg(37)*(vx*t-25)

    h=vy*t-4.05t²
    tg(37)*(vx*t-25)=vy*t-4.905t²

    Like this?
     
  11. Jan 12, 2014 #10
    d=vx*t-25 correct
    h=vy*t-4.05t² correct


    tg(37)*(vx*t-25)=vy*t-4.905t² now solve for t and you are basically done. Yes, its a quadratic.



    No
     
  12. Jan 12, 2014 #11
    You are right, when I do that I get 1.06 seconds and when I add it to 1.96 I get 3.02 seconds.

    And doing this gets me the same result, 3.02 seconds.

    Thanks for the help.
     
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