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Projectile motion on an inclined plane

  1. Jan 26, 2017 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown with initial speed v0 up an inclined plane. The inclined plane makes an angle of π/6 above the horizontal line and the ball is launched at an angle θ above the inclined plane. No air resistance in this problem.
    (a) How long does the ball stay in the air?
    (b) At what angle θ should the ball be launched in order to fall back on the plane normal to the inclined surface?

    2. Relevant equations
    Newton's laws and kinematics equations?

    3. The attempt at a solution
    I tried making a rotated reference frame (expressed as a matrix transformation below) centered at the launch point where $$ \begin{pmatrix} x' \\ y'
    \end{pmatrix} = \begin{pmatrix} \cos(π/6) & \sin(π/6) \\ -\sin(π/6) & \cos(π/6) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\
    \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac {\sqrt{3}} 2 & \frac 1 2 \\ - \frac 1 2 & \frac {\sqrt{3}} 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}
    and since the force of gravity is ##\vec {\mathbf F} = -mg ~ \hat y##, then in the new reference frame $$ F_x' = \frac {\sqrt{3}} 2 F_x + \frac 1 2 F_y = - \frac 1 2 mg \\ F_y' = - \frac 1 2 F_x + \frac {\sqrt{3}} 2 F_y = - \frac {\sqrt{3}} 2 mg$$
    and acceleration ##a_x'## and ##a_y'## will just be the above two forces with mass divided out.

    From here on I'm a bit stuck. As for part (a), I want to use the kinematic equations with my given initial velocity. At first I thought to solve for the time it takes to reach the peak of its trajectory and then multiplying by 2, but I realized that would be the time it takes to hit the ground if the inclined plane weren't there. My next guess was solving for the time when the horizontal velocity reached zero since in this reference frame there's a negative horizontal acceleration, but I'm not sure that would be the exact time that the ball hits the inclined plane.

    Any help would be appreciated!
  2. jcsd
  3. Jan 26, 2017 #2


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    OK. That's a very formal way to do it, but it works! You could also just project the acceleration due to gravity along your x' and y' axes.


    Your idea that the x'-component of velocity should be zero at impact might be right for part (b). But your reason for stating that doesn't make sense to me. Keep in mind that for part (b) you want the ball to hit the incline perpendicularly to the incline.

    For part (a), think about the coordinates of the point of impact in your primed coordinate system.
  4. Jan 26, 2017 #3
    Thanks for your input! My professor wants us to practice rotated reference frames using matrix transformations, hence this homework problem.

    I think I figured out part (a). I just used the equation ##y' = v_{0,y}' t + \frac 1 2 a_y' t^2## and set it equal to 0, then solve for ##t##. I'm still working out part (b).
    Last edited: Jan 26, 2017
  5. Jan 26, 2017 #4


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    Good. (Forgot to type the square on the t in the last term?)
  6. Jan 26, 2017 #5
    Ah, right I did. Thank you!

    I believe I have the second part too now. I'll just set ##\vec {\mathbf r}' \cdot \hat {\mathbf y} = 0## because the dot product is zero when the vectors are orthogonal. The x-component drops out, and the position ##y'## will be from the equation I just used, evaluated at the time that I solved for.
  7. Jan 26, 2017 #6


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    So, you are saying that the position vector is perpendicular to the y'-axis at impact? (I'm interpreting ##\vec {\mathbf r}'## to be the position vector in the primed frame with origin at the launch point.) Wouldn't that always be true for any impact location on the plane (even if the ball does not hit perpendicularly to the plane)?

    But you already know that ##y'## will be zero at impact.

    Try to formulate the idea that the ball is moving perpendicular to the inclined surface at impact.
  8. Jan 26, 2017 #7
    Oops, yeah I realized that wouldn't work after trying it. Do you think I could set ##\vec {\mathbf v}' \cdot \hat {\mathbf x}' = 0## instead? Evaluating the velocity at the time I solved for, the vector points "down" in the coordinate system, which is perpendicular to the coordinate's unit vector x. Then I solve for θ.
  9. Jan 26, 2017 #8


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    Yes, sounds like a good plan.
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