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How can i launch a golf ball 15 m?

  1. Sep 27, 2007 #1
    hey guys, my classmate and i have to design a device that will launch a golf ball 15m, using either a spring (which is what we're going to use) or elastic
    we need to find the value of the spring constant(we can do that), an expression to determine the initial velocity of the ball, an equation to determine the angle required to hit the target 15m away (using the initial velocity), and lastly, an equation where we can sub in how far back we have to pull the spring and the angle we have to aim the device to launch the ball to any random displacement (the teacher will tell us on the day of)

    1. The problem statement, all variables and given/known data
    dx = horizontal displacement
    dy = vertical displacement
    ay = vertical acceleration (aka. gravity)
    t = time/change in time
    mass of ball = 45.7g
    dx = 15.0m
    dy = 0.00m (since the device will be put on the ground, the ball will be launched and will land back on the ground, therefore the vertical displacement will be 0m)
    ay = -9.81m/s^2

    2. Relevant equations
    dx = vx(initial)*t, vx(initial) = 15.0m/t

    dy = vy(initial)*t + 0.5(ay)(t^2), 0.00m = vy(initial)*t + 0.5(-9.81m/s^2)(t^2)
    vy(initial) = 4.905m/s^2(t)

    tan(angle) = opposite/adjacent = vy(final)/vx(final)

    vx(final) = vx(initial)
    vy(final) = vy(initial) + a*t = vy(initial) + (-9.81m/s^2)*t

    3. The attempt at a solution
    this is where we're stuck... we're not sure how to find the time it takes for the ball to travel a distance of 15.0m, therefore we don't know how to find the initial velocity, nor do we know how to figure out the angle

    sorry for the confusing equations and such

    thanks for the help! :)
     
  2. jcsd
  3. Sep 27, 2007 #2

    CompuChip

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    Homework Helper

    So, suppose you have an initial velocity [itex]v_0[/itex] at an angle [itex]\theta[/itex] from the horizontal. First express the horizontal and vertical component of the velocity in [itex]\theta[/itex] (something with the magnitude [itex]v_0[/itex] and a sine or cosine). Now the horizontal velocity does not change. The vertical velocity does: there is an acceleration -g due to gravity. So you should find a formula that expresses the vertical distance in g and the time t, solve for the time t' at which the ball hits the ground and plug this number into the formula for horizontal distance (which is just [itex]s_\mathrm{hor}(t) = v_\mathrm{hor} \cdot t[/itex] to get the traveled distance. This will link [itex]v_0[/itex] to s(t') and allow you to find out which [itex]v_0[/itex] you need to get s(t') = 15 m.

    It sounds complicated, but just try it (step by step) and post what you got. And don't be afraid to do some algebra (solving quadratic equations and the like)
     
    Last edited: Sep 27, 2007
  4. Sep 29, 2007 #3
    thanks for the reply, CompuChip! and sorry for the delayed reply... my friend and i have been so busy with trying to build the actual device. :(

    we just had a question... what exactly is shor(t)? thanks! :)
     
  5. Sep 29, 2007 #4

    learningphysics

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    shor(t) is horizontal displacement.
     
  6. Feb 18, 2008 #5
    i have a similar physics project, and wanted to know what kind of device you guys ended up building and how everything worked out for you guys
     
  7. Feb 18, 2008 #6
    to be honest, we gave up after a while... and so we just bought a slingshot :)
    sorry i couldn't be of more help.
     
  8. Feb 18, 2008 #7

    DaveC426913

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    Gold Member

    I know a girl in this Texan peeler joint who can do this trick. I think I've got her number...
     
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