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Motion - Projectile

  1. Sep 18, 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 30 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
    [PLAIN]http://img36.imageshack.us/img36/8817/bookscanstation20110918.jpg [Broken]


    2. Relevant equations

    v^2 = v_0^2 + 2 a delta x
    v = v_0 + a t
    x = x_0 + v_0 t + 1/2 a t^2

    3. The attempt at a solution

    I set point were the ball is originally to be represented by the Cartesian coordinates (0,0) and t = 0. I choose to the right of this point to be positive, above this point to be positive, and below this point to be negative.

    I solved for the maximum height of the ball
    v^2 = v_0^2 + 2a delta y
    delta y = (v^2 - v_0^2)/(2a)
    Used the fact that the velocity at the top of the ball is zero, the initial velocity in the y direction is 10 sin(30) = 5 m/s, and the acceleration due to gravity is about 9.8 m/s^2 downwards
    delta y = - 5^2/(2*-9.8) = 1.276 meters

    I solved for the time it takes for the ball to get to this point
    v = v_0 + at
    t = (v - v_0)/a
    t = -5/-9.8 = .510 seconds

    I than solved for the horizontal displacement at this point
    If we ignore any type of acceleration in the horizontal direction
    x = v_0 t
    v_0 in the x direction is 10 cos(30) = 5sqrt(3)
    x = 5 sqrt(3) t
    x = 5 sqrt(3) .510 = 4.417 meters

    I than solved for how far below the ground is directly below this point x = 4.417 level to the point that can be represented to describe the point right before the ground starts slanting downwards (0,-1) using trigonometric functions
    4.147 tan(45) = 4.147 meters

    I than concluded that the equation that can represent the vertical fall at any point from y=-1 to be
    y = 5sqrt(3)t

    Using the general equation for displacement for the vertical direction
    y = y_0 + v_0 t + 1/2 a t^2
    and setting y (the point that represents the final height of the ball) to be the opposite of 4.417 + 1 + 5 sqrt(3) t
    or simple
    -(5.417 + 5sqrt(3) t)
    I than set this equal to the right hand side of the equation (the initial velocity in the vertical direction of the ball at the top of it's trajectory is zero) and using the point 1.276 (the point that can be used to describe the height of the ball at the top of it's trajectory) I got
    -(5.417 + 5sqrt(3) t) = 1.276 - 4.9 t^2

    I than used the solve feature on my calculator and got that
    t = 2.349 seconds

    I than plugged this into the equation that represents the horizontal displacement of the ball
    x = 5 sqrt(3) 2.349
    x = 20.343 meters

    The answer is 24.85 meters and I don't see what I'm doing wrong. Thanks for any help that you can provide.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 18, 2011 #2
    Hi,
    If you don't mind, in spite of your very engrossing work, I'd define how I would've solved the problem:
    The slope with the 45 deg. incline, has a function... Any interaction of the projectile with it, depends on the collision with it, or in other words, an intersection of their respective coordinates.
    Lets use a different coordinate system than your own, if you don't mind(though it doesn't matter), Point O(0,0) is at the start of the slope with 45 degs.
    It has the equation [itex] y = -\arctan{\alpha}\cdot x, \alpha = {45}^o , y = -x [/itex]
    Suppose the particle falls on the slope at some point [itex] x_f [/itex]; Then, on the slope, it has a [itex] y_f = -x_f [/itex] but also, since the particle's y tranjectory is delineated by:
    [itex]\Large y = y0 + v_i\sin{\beta}t-\frac{gt^2}{2}, \beta = {30}^o [/itex], and the reaching the x_f is conditioned by the particle's velocity x component which prompts:
    [itex]\Large x_f = v_i\cos{\beta}t_f, t_f = \frac{x_f}{v_i\cos{\beta}} [/itex].
    Putting it all together, by plugging it in y_f, on both sides:
    [itex]
    \Large
    -x_f = y_0 + x_f\tan{\beta} - \frac{g}{2}\frac{{x_f}^2}{{v_i}^2{\cos^2{\beta}}}
    [/itex]
    Which is a simple quadratic equation, yielding two roots, the negative one can be thrown out, and the positive one is your answer...
    Hope that works for you,
    Daniel
     
    Last edited: Sep 18, 2011
  4. Sep 18, 2011 #3

    PeterO

    User Avatar
    Homework Helper

    With your definition of origin and directions, the equation of the landing hill is y = -x -1. If you can find the equation of the path of the projectile you could solve simultaneous equations to get the answer.
     
    Last edited by a moderator: May 5, 2017
  5. Sep 18, 2011 #4
    Thanks for the responses.
    - x tan(45) = y
    - x = y

    y = y_0 + v_0 t + 1/2 a t^2
    v_0 = 10 sin(30) = 5
    a = -9.8
    y_0 = 1
    y = 1 + 5 t - 4.9 t^2

    x = v_0 t
    v_0 = 10 cos (30) = 5 sqrt(3)
    x = 5 sqrt(3) t

    t = x/(5 sqrt(3) )

    y = 1 + 5 t - 4.9 t^2
    - x = 1 + 5 t - 4.9 t^2
    - x = 1 + 5(x /(5 sqrt(3) ) ) - 4.9 (x/(5 sqrt(3) ) )^2
    - x = 1 + x/sqrt(3) - (4.9 x^2)/75

    When I solve I get 24.76 not 24.85 am I doing something wrong?
     
  6. Sep 18, 2011 #5
    No, in fact you're quite right... this mainly depends on your "g".
    Try using g=9.766, that should get you closer.
    This is a mere rounding error, and should be ignored, you've done everything properly...
     
  7. Sep 18, 2011 #6
    Ok, thank you very much
     
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