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Projectile motion on slope

  1. Sep 16, 2010 #1
    A golfer drives a ball horizontally with
    initial velocity v = (50m/s , 0) from a tee at the
    origin, down a 20deg below-horizontal slope as
    illustrated above.

    A. How far from the tee measured along
    the slope does the ball land on the slope?
    B. With what speed does it land?


    a link to the page in the book with the problem: http://s861.photobucket.com/albums/ab174/alkaline262/?action=view&current=chapter2question.jpg



    x(t) = x0 + v0x*t + 1/2*a*t^2
    y(t) = y0 + v0y*t + 1/2*a*t^2





    x(t) = v0*t
    y(t) = -1/2*g*t^2

    solving x(t) for t:

    t = (x/v0)

    inserting in y equation fro trajectory:

    y(x) = -1/2*g*(x/v0)^2

    i will call the projectile range R, as it would be along the x axis. to compensate for the slope of the hill i will now substitute into the trajectory equation:

    R*sin(θ) = -1/2*g*([R^2*cos(θ)^2]/v0^2)

    solving for R i get (sin(θ)*-2*v0^2*)/(cos^2*g) = R

    inserting v0=50 and θ=-20deg i get the range to be 197

    my prof says the range is 185, which i would have gotten i the cos of the denominator was not squared, where is my error here?

    I am still working on the second part, but need to clear this up first. Any help would be greatly appreciated


     
  2. jcsd
  3. Sep 16, 2010 #2

    kuruman

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    Consider the above equation. You also know that tan20o = -y/x. So ...
     
  4. Sep 16, 2010 #3
    xtan(-20) = -1/2*g(x^2/v0^2)

    x = [2*v0^2*tan(-20)]/-g = 185

    to find the intersection, when the y vales are equal. this comes out to 185, but i dont understand if my math above(when i did it the long way) is incorrect. how did i make a mistake? was it in the math, or was that just altogether incorrect? Thanks for your help so far.
     
  5. Sep 16, 2010 #4

    kuruman

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    I am not sure where you got this

    R*sin(θ) = -1/2*g*([R^2*cos(θ)^2]/v0^2)

    That's probably where you went wrong.
     
  6. Sep 16, 2010 #5
    all right, ill just forget it then. thanks for your help; it was substantial
     
  7. Sep 16, 2010 #6
    so far we have said

    x= [2*v0^2*tan(20)]/g

    since tan = -y/x

    then y = [-tan(20)^2*2*v0^2]/g

    right?

    if that is correct, then can i differentiate these two to get the velocities (vx,vy) then plug that into the equation for finding speed of impact :

    IvI = √ (vx^2 + vy^2)

    if that is right, then it is completely solved!
     
  8. Sep 16, 2010 #7

    kuruman

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    Right. But x in the above equation is not a function. It is the range R, a constant. Same for y, it stands for the vertical drop when the horizontal distance is R.

    Differentiate what two and with respect to what? The horizontal velocity vx is constant and equal to 50 m/s. You can find the vertical velocity vy if you first find the time of flight and then use vy = gt. Having the two components, you can find the impact angle.
     
  9. Sep 16, 2010 #8
    :blushing: welp, back to the drawing board.

    185*cos(-20)=v0*t

    [185*cos(-20)]/50 = t

    3.5s = t

    now i believe i can solve for the speed of impact.

    given by the equation √(vx^2 + vy^2)

    speed of impact = 60.6 m/s

    Hopefully that is it, i see what you mean about the derivatives and constants. Thank you for your help
     
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