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Calculation to launch a golf ball

  1. Mar 3, 2015 #1
    Hi,

    I need your help. Is my calculation correct?


    1. The problem statement, all variables and given/known data

    We have to build a launcher that will be able to reach a target that is place anywhere in a prism 3m X 3m X 1m. The launcher have to used a golf ball and gravity as source of energy.

    I am trying to calculate what we need as energy. :

    -We are thinking about using a pendulum.

    -We want it to hit the ball in a tube that we could control to adjust the angle of launching.​

    We know :

    -that we have to used 45 degrees to maximize our shot.

    -that the ball golf as a coefficient of restitution of 0,83 and a mass of 45,9 g.

    -that the longest distance we will have to reach will be at 1,5 m to the left or the right, 1 m from the ground and at 3 m from the launcher.​

    We think of using a mass of 100g for our pendulum.

    2. Relevant equations
    By Pythagoras's theorem, we can calculate the distance we want to reach.

    [itex]c^2 = a^2 + b^2[/itex]

    Using cinematics equations, I have this :

    [itex]x=V_ {xo} * t[/itex]

    [itex]y=V_ {yo}* t - \frac{1}{2} g t^2[/itex]

    The equation for a inelastic collision is :

    [itex]V_ {2}=\frac{C_r m_1 V_1 + m_1 V_1}{m_1 + m_2} [/itex]

    For the pendulum, we have the equations of the Law of the Conservation of energy

    [itex] \frac{1}{2}mV^2= m g h[/itex]

    3. The attempt at a solution
    If we have an angle of 45 degrees, we know that the speed in x and in y will be the same and will be equal to :

    [itex]V_ {xo}= V_ {yo} = V sin 45[/itex]

    By Pythagora's, we know that we need to reach 3,35 m in X and 1 m in Y.

    With the equations of cinematics, I've got :

    [itex]V=\frac{x}{\sqrt{(\frac{2(x-y)}{g})}}[/itex]

    and that give me [itex]V= 3,42 m/s[/itex]

    After that, as I know that the ball golf as a coefficient of restitution of [itex]C_r = 0.83[/itex], I have calculate that I need a speed of [itex]V= 4.2896 m/s[/itex] with the equation of inelastic collision.

    Finally, with the equations of the pendulum, I have calculate that I have to have a height of [itex]h = 0.937 m[/itex].


    Is that correct?
    Thank you!!
    (Sorry if I have made any english errors)
     
  2. jcsd
  3. Mar 3, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    Where does the "2" come from in the 2(x - y) term inside square root? Can you show your derivation?

    Note that for 45°, sin(45) = cos(45) = 1/√2.
     
  4. Mar 3, 2015 #3
    It came from the isolation:
    1- [itex]x=V_ {xo} * t[/itex]

    2- [itex]y=V_ {yo}* t - \frac{1}{2} g t^2[/itex]

    In the second equation, I replace t by [itex]t=\frac{x}{V_ {xo}}[/itex] and I replace [itex]V_ {xo} =V_ {yo}=Vsin45[/itex]

    I now have :
    [itex]y=\frac{Vsin45}{Vsin45}*x - \frac{1}{2} g (\frac{x}{Vsin45})^2[/itex]
    And I isolate
    [itex]y=x - \frac{1}{2} g (\frac{x}{Vsin45})^2[/itex]
    [itex]\frac{1}{2} g (\frac{x}{Vsin45})^2 = x-y[/itex]
    [itex](\frac{x}{Vsin45})^2 = \frac{2(x-y)}{g}[/itex]
    [itex]\frac{x}{Vsin45}) = \sqrt{\frac{2(x-y)}{g}}[/itex]
    [itex]Vsin45 = \frac{x}{\sqrt{\frac{2(x-y)}{g}}}[/itex]
    And now I realised I have forgotten the sin 45 so :
    [itex]V = \frac{x}{\sqrt{\frac{2(x-y)}{g}}} * \frac{1}{sin45}[/itex]​
     
  5. Mar 3, 2015 #4

    gneill

    User Avatar

    Staff: Mentor

    Okay. And sin(45) is 1/√2, so...
     
  6. Mar 3, 2015 #5
    So I would need 6,84 m/s.
    Then, because of the inelastic collision, I would have to have a speed of 8,57 m/s for my pendulum and a height of 3,7m??
     
  7. Mar 3, 2015 #6

    gneill

    User Avatar

    Staff: Mentor

    That looks good.
    Hmm. I did the calculation quickly, but I didn't see a value that large: Something between 5 and 6 m/s.
     
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