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Motion Question

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data

    A ball is launched from a cliff 1 meter high. At the bottom of the cliff, the ground is not flat, it slants downward at a 45 degree angle as show. The ball's initial; velocity is 10 m/s and makes a 20 degree angle with the horizontal. Find x, the horizontal displacement of the ball from the edge of the cliff when it lands.

    Ans: 24.85 m

    Please see the picture that describes the problem in the attachments

    2. Relevant equations

    3. The attempt at a solution

    When I tried solving this problem I got about 23.32 meters so I guess I'm doing something wrong.

    What I did was solve for the equation that describes the position in the vertical direction as a function of time and got
    y = -4.9*t^2 + 5t
    I set this equal to negative one to find the time at which the ball was one meter below were it started (I set the coordinate system to be (0,0) were the ball originally starts), and got about 1.1916656. I then plugged this value into the equation that gives the displacement in the horizontal direction
    x = 5sqrt(3)*t
    x = 5sqrt(3)*1.1916656
    x is about 10.32012682
    I then solved for how far below the ground was from this point using trigonometry were
    10.32012682 tan(pi/4) = 10.32012682, 10.32012682 = 10.32012682
    from this I concluded that the point directly below the ball when the ball is exactly one meter below the point from were it started can be represented by the point (10.32012682,-11.32012682)
    I then concluded that the equation that can describe the vertical position of the ground were it's slanted can be represented by the equation y(x)=-x-1
    I then got rid of the parameter t in my equation that describes the vertical position of the ball by solving my equation that represents the horizontal position of the ball for t
    x(t) = 5sqrt(3)t
    t = x/(5sqrt(3))
    y(t) = -4.9*t^2+5t
    y(x) = (-4.9x^2)/25.3 + (5x)/(5sqrt(3)) = -(49x^2)/750 + (sqrt(3)x)3
    I then set my equation that describes the height of the ball as a function of horizontal location to the equation that represents the height of the ground as a function of horizontal location
    y = -(49x^2)/750 + (sqrt(3)x)3
    y = - x - 1
    -(49x^2)/750 + (sqrt(3)x)3 = - x - 1
    I got that x is about 24.76 which is off from the answer on the sheet by about .9. Am I doing something wrong?

    Thanks for any help anyone can provide me!

    The picture
    http://img36.imageshack.us/img36/8817/bookscanstation20110918.jpg [Broken]
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 17, 2011 #2
    Having a bit of trouble understanding what you did, so if i type some of what you wrote, i'm sorry. I would also say you might have gotten help quicker in the normal physics homework section seeing as this is just basic kinematics.

    The first thing you would like to do is to find the time it takes for the ball to reach it maximum height then find the distance in the x direction at that point. You can also find the height from the ramp at this point knowing it's declination and starting height from the ramp.

    I'd like to say that it would be best to solve this analytically ( so if i used itex 20 degrees woudl be theta and 45 would be phi)

    vy = vsin20
    vx = vcos20 which is constant the entire till it bounces off the other ramp

    vyf = 0 = vyi + at t = vyi/g calculate the height and distance

    dis y = ((vyi)^2)/g + 1/2g(vyi/g)^2

    dis x = vix(t) = vix(vyi/g)

    now from the ramp would be dis y + 1 + dis x tan(45) = total initial height before falling

    the distance x on the declination is changing at vcos20, so y is changing by vcos20

    now you want to calculate the time it takes for the ball to reach the ramp

    You can consider that the way the ball is moving it is slowing down because it is getting closer to the ramp with time. Sorry i've ran out of time, but it should work.

    Just a note you've changed the initial ramp conditions from 30 to 20 degrees
  4. Sep 17, 2011 #3
    Hm thanks for the advice and ya I wasn't sure were to put it lol
    so it takes .510 seconds for the ball to reach the top of it's bath
    at this point in time it's 1.276 meters above were it started
    at this point in time it's 4.417 meters horizontal from were it started
    at this point it's 6.693 meters above the ground

    So the point below the ball right when it's at the top of it's path can be described by the point (4.417m, -5.417)
    y = y_0 + v_0 t + 1/2 a t^2
    y is changing with time by 5sqrt(3) t [sorry for getting the angle mixed up] cos(30) = sqrt(3)/2
    -(5.417 + 5sqrt(3)t ) = 1.2716 - 4.9 t^2
    I get t is about 2 seconds

    plug back into

    x = 5sqrt(3) t
    x = 5sqrt(3)2 is about 17.321

    i guess I'm doing something wrong still
  5. Sep 18, 2011 #4
    I'll post this in the easier physics section or whatever, I keep on getting the wrong answer still for some reason
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