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Homework Help: Projectile fired down a slope

  1. Apr 3, 2015 #1
    1. The problem statement, all variables and given/known data
    An archer standing on a 15 degree slope shoots an arrow 20 degrees above the horizontal. How far down the slope does the arrow hit if it is shot with a speed of 50m/s from 1.75m above the ground?

    2. Relevant equations

    ##y = x##tan##\theta## - ##\frac{gx^2}{2v^2 cos^2\theta}##

    3. The attempt at a solution
    I took the slope the archer is standing on to be a line through the origin of the x-y coordinate system, thus intersecting the trajectory of the arrow at two points, one being (0, 0). The slope this line is then

    ##y = -(tan15^o)x##

    Setting the equations equal to each other and punching in values I obtained

    ##-(tan15^o)x = xtan20^o - \frac{9.8x^2}{5000cos^2 20^o}##

    ##x = \frac{(tan15^o + tan20^o)(5000cos^2 20^o)}{9.8}##

    with x = 284.7 and y = -76.28 giving a final distance of 294.7m. The actual answer is 297m according to the book. I've been over this calculation a few times and it is clear that another set of eyes is required to shed some light on this conundrum. Please assist.
  2. jcsd
  3. Apr 3, 2015 #2


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    Did you take care of the fact that the arrow is shot from 1.75 m above the ground?
  4. Apr 3, 2015 #3
    initial upward velocity = sin(20)*50.
    initial y = 1.75

    yslope= mx+b
    yslope= tan(-15)*x+0

    tan(-15 deg)*cos(20 deg)*50*t=-.5*9.8*t^2+sin(20 deg)*50*t+1.75


    I tried. It's possible the book has a typo.
  5. Apr 3, 2015 #4

    Doc Al

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    You don't seem to have taken into account the initial height of 1.75 m.

    (TSny beat me to it.)
  6. Apr 3, 2015 #5
    It seems I did not, and shifting the curve upward by 1.75m does indeed solve the problem.
    Something else though. Initially I thought about rotating the system so that the ground is level and the launch angle is 35 degrees. Am I correct that this would not work? If one considers say, the same 15 degree slope and a launch angle of 75 degrees, rotating would simply send the arrow straight up, not so?
  7. Apr 3, 2015 #6


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    Right. It would not work unless you also rotated the direction of the acceleration of gravity.

    That's right (assuming gravity still acts vertically)
  8. Apr 3, 2015 #7
    Of course. I didn't consider that, but I can see now that going about it that way wouldn't be simpler at all, seeing that compensating for the direction of gravitational acceleration would result in a horizontal component thereof as well.
    Thanks for the help.
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