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A perplexing projectile problem

  1. Oct 9, 2005 #1
    hi, i guess this is my first post. and it will be somewhat of a doozy. i hope someone out there can help. the problem goes as follows.

    a punter kicks a football such that it just clears a 10.0m high barrier 30.0m away. it continues another 10.0m to the ground. with what velocity (speed and angle) did the punter kick the ball?

    please help. this is horribly confusing to me.
  2. jcsd
  3. Oct 9, 2005 #2


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    This is a pretty challenging question for starters. I assume you're aware of the kinematics equations. The trick always is to find 3 variables out of 5 (time, acceleration, initial vel, final vel, displacement), then you can solve for the other two. However in this question, there are 2 components, x and y, so we'll need to break it up. Assume ground level is 0,0.

    Working in the x direction, clues given are that displacement is 20 meters. Acceleration is zero (as per usual in the horizontal), initial velocity is the unknown parameter.

    Working in the y direction, clues given are displacement of interest is 30 meters. Acceleration is -9.8m/s/s (as per usual in vertical), initial velocity is unknown parameter (same from x).

    In y: Vinitial x sin (angle)
    In x: Vinitial x cos (angle)

    Your task, solve for Vinitial and angle. Don't forget to draw the question out. It helps tons.
  4. Oct 9, 2005 #3


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    amoretti, please read the sticky thread (linked to in my signature below) for help with homework problems.
  5. Oct 9, 2005 #4
    okay, i'm working it out as best i can, but nothing seems to fit. i keep getting stup with useless information, and my limitted knowledge of trig identities (i'm in 11th grade) does not help. btw, sorry, goku, if it seemed like i was asking for an answer. i'm simply looking for a means.

    i've found three sets of (seemingly) pertinent information:

    ending where the ball lands:
    vertical: vi=vsinø, acc=-9.80m/s^2, time=?, disp=0, vf=-vsinø
    horizontal: vave=vcosø, acc=0, time=?, disp=40.0m

    ending at the 10.0m high barrier:
    vertical: vi=vsinø, acc=-9.80m/s^2, time=?, disp=10.0m, vf=?
    horizontal: vave=vcosø, acc=0, time=?, disp=30.0m

    ending at the maximum height the ball reaches:
    vertical: vi=vsinø, acc=-9.80m/^2, time=?, disp=?, vf=0
    horizontal: vave=vcosø, acc=0, time=?, disp=20.0m

    any hints on where i should start, since all of these seem equally reasonable?
  6. Oct 9, 2005 #5


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    Here's a hint. We know the range is 40 m. Neglecting air drag we also know the trajectory is symmetric so the height of the football is 10 m when it has travelled 10 m horizontally from the start position.
  7. Oct 9, 2005 #6
    yea, but i have no idea what to solve for, regardless if that.
  8. Oct 9, 2005 #7


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    As mezarashi said, this is a pretty challanging question. I've done it and and there was a fair amount of work and a lot of manipulation involved with the expressions.

    You've got three sets of information. You can forget about the last one - max height reached, although it sort of does get involved later on, but indirectly.

    Set up eqns of motion for distance moved in terms of velocity and time. In both the x- and y-directions.

    Using Tide's hint, you have two locations you know the x- and y-positions of -- the final/end point, (x,y) = (40,0) and the top of the wall, (x,y) = (30,10) or, (10,10), by symmetry.

    Apply the two eqns of motion to those positions, then do a lot of manipulating.

    You will have to solve for both v and θ. I found it easier to get θ first.
  9. Oct 9, 2005 #8
    thanks, i'll give this a shot.
  10. Oct 9, 2005 #9
    alright, i've been working on this for a while now, and i just need an answer check. fermat, did you get 53.1degrees for the angle?
  11. Oct 9, 2005 #10


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    Hi, I got tanθ = 4/3.

    I never converted it to degrees. I recognised it as coming from a 3-4-5 triangle and just wrote down sinθ = 4/5, cos = 3/5.
    Last edited: Oct 9, 2005
  12. Oct 9, 2005 #11


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    Dearly Missed

    A tangent value equal to 4/3 sounds suspiciously close to 53.1 degrees.
  13. Oct 9, 2005 #12


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    more text to make up the 6-figure limit
  14. Oct 9, 2005 #13
    yay, i got it, thank you all so much.
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