Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Projectile Motion Off a Slope.

  1. Jan 15, 2007 #1
    1. The problem statement, all variables and given/known data
    A ball is kicked off of a hill with an initial horizontal velocity of 10 m/sec.
    The pitch of the hill from where the ball is thrown is -20°. Determine how far the ball lands from its original position.

    Not given, but I'm acting under the assumption that gravity = -9.81m/sec²

    This is how I see the problem,
    http://img78.imageshack.us/img78/6411/projectileeu0.gif [Broken]

    2. Relevant equations
    a = v/t
    d = vo*t + ½at²

    3. The attempt at a solution

    I would love to say that;
    a = v/t
    9.81 = 10/t
    t = 1.02sec

    Then just split and solve so that
    d(x) = 10.2, d(y) = -5.1
    √(dx² + dy²) = 11.4m

    But I can't do that because there is no y velocity so, solving for the only constant.
    d = vo*t + ½at²
    Split into x,y.

    d(y) = -4.905 t²
    t = √(d(y)/- 4.905)

    d(x) = vo(x)*t
    t = d(x)/10

    Setting them equal,
    d(x)/10 = √(d(y)/- 4.905)

    dx = 10√(dy/- 4.905)
    dy = - 4.905(dx/10)²

    To find the distance from start use law cosines, or in this case pythag.

    d = √[{10√(dy/- 4.905)}² + {- 4.905(dx/10)²}²]

    While this is nice, it isn't getting me any closer to the answer. I need to solve for time somehow...
    So... If anyone can give me a nudge it would be extremely grateful. :blushing:
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 15, 2007 #2


    User Avatar

    Staff: Mentor

    You are close, but don't set dx and dy equal. Set them in the ratio of the slope....
  4. Jan 15, 2007 #3
    By setting them in ratio do you mean,

    atan[{√(d(y)/- 4.905) }/{d(x)/10}] = 20°

    tan 20° = .3640

    .364(d(x)/10) = √(d(y)/- 4.905)

    Which means I am still left with two variables and one equation, or, do you mean.

    dx : dy = cos(20°) : sin(20°)

    dy = 2.747 dx
    dx = 0.364 dy

    which is nice, but once again I fail to see how it helps :frown:
  5. Jan 15, 2007 #4


    User Avatar
    Homework Helper

    It seems to me you are overcomplicating things.

    Take the equations of displacement in the x and y direction, and let x0 and y0 be the displacements at some point in time t0. You have x0^2 + y0^2 = d^2, where d is the displacement you need to find. You know the slope of the hill, so you can combine that with the kinematic equations of displacement and solve for t0, and then easily obtain your displacements x0 and y0, i.e. d.
  6. Jan 16, 2007 #5


    User Avatar

    Staff: Mentor

    Hi sroy2,

    By using the slope, I meant start with your equations for the x and y motion:

    and use the fact that the x and y displacements would have to be in the ratio of the slope when the projectile makes final contact.

    But I like radou's simpler method better.
  7. Jan 19, 2007 #6
    answer to the football kicked on a slope question

    okay this is a straightforward question (you don't need to solve for time anywhere in this !!)

    for my equations I am using the following notations

    a = acceleration u = initial velocity v = final velocity s = displacement
    t = time (taking acceleration due to gravity as 9.8ms-2)

    try and ignore the slope and just look at this ball as a horizontally projected projectile

    by considering the motion horizontally

    speed = distance/time (since acceleration is always zero horizontally)

    lets call this horizontal distance y

    then 10 = y/t therefore t = y/10

    by considering the motion vertically
    lets call the vertical distance moved down by the ball during flight X

    therefore vertically, (we'll come back to this calculation later)
    a = 9.8 ms-2
    u = 0
    s = X
    t = y/10

    now we must express y in terms of X in order to use the constant acceleration formulae

    now imagine the slope.......the distance moved VERTICALLY DOWN is X and the distance moved HORIZONTALLY ACROSS is y. Now lets call the distance moved along the GROUND d ......we have a triangle with hypotenuse d and sides X and y...........and we have a value for the angle, 20 degrees

    so by trigonometry, the side y in terms of x would be x/tan20

    now going back to our equation for horizontal motion

    t = y/10 we can now say t = (X /tan 20) /10

    i.e. t = X/ (10tan20) hoooray we got rid of the y term !


    considering motion vertically

    a = 9.8
    u = 0
    s =X
    t = X/(10tan20)

    using the following constant acceleration formula

    s = ut + 1/2 at^2
    x = 0 x t + 1/2 x 9.8 x (X/10tan20)^2
    x = (4.9X^2) / (10tan20)^2
    (10tan20)^2 = 4.9X

    X = 2.704 m well now that we have the value of X and the angle (20 degrees), we want to the know the distance moved along the ground d

    so using trigonometry sin 20 = X / d
    sin20 = 2.7/d
    d = 2.7/sin20
    d = 7.91
    Last edited: Jan 20, 2007
  8. Jan 24, 2007 #7
    Chromium Blade,

    I solved a similar problem using your method, but my answer was off by a factor of 2 (33.15 ft instead of 66.3 ft)

    My problem: A ball bounces on a 30 degree inclined plane such that it rebounds perpendicular to the incline with a velocity of 40 ft/s. Determine distance R (length of plane) to where it strikes the plane.

    I set up my coordinates so that horizontal is X and vertical is Y (opposite of yours; so keep that in mind).

    Vx = 40sin30 = 20ft/s
    Vy = 40cos30 = 34.64 ft/s

    20 = x/t, therefore, t = x/20; x = y/tan30

    So, t = y/20tan30 = 0.0866y

    y = Vy*t + (1/2)*a*t^2

    y = (34.64)(0.0866y) + (1/2)(-32.2)(0.0075y^2)

    y = 3y - 0.1207y^2

    0.1207y^2 - 2y = 0

    Solve for y:

    y = 16.57


    R = 16.57/sin30 = 33.14 ft

    Whereas the answer is 66.3 ft.

    Plugging y to solve t, t = 1.435 s, whereas the time should be 2.87 s.

    I know i'm probably doing something really stupid, but I can't seem to find where i'm off by a factor of 2.

    Any help would be greatly appreciated.

  9. Jan 25, 2007 #8
    Hi David

    It was all going great until your vertical consideration of motion !

    Quite simply, if the ball is going to be landing BELOW the LEVEL of projection, its OVERALL displacement is negative !! In my other example, displacement was positive because the ball was horizontally projected, but in your question, it is projected at an angle to the horizontal, so we must make every value vertically above the projection level positive, and every value below the projection level negative.

    so displacement = -y

    so s = ut + 1/2 at^2 your formula is 100% correct !

    only s= -y in this case, notice I've written this in my working out for an earlier post. It's negative displacement because the ball is ending up BELOW ITS STARTING LEVEL. Now try to solve that quadratic. Come back to me if you have any problems, best of luck!

    Chromium blade
    Last edited: Jan 25, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook