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Homework Help: Did I solve this right?

  1. Nov 5, 2012 #1
    This problem was given to me on a test today. I tried my best to figure it out but I'm still not sure, I did find a solution, I am just not sure if the solution is correct.

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

    A long jumper jumps with an angle of 23 degrees and lands 8.59m away from the jump spot. What is the magnitude of the initial velocity of the jumper?

    The feet land in the same place as the body (the body is to be treated as an object)
    Horizontal motion is constant
    Given: (small "v" is speed, large "v" is velocity)
    x - components
    a = 0
    Vi = vcos23
    d = 8.59m
    t = ?

    y - components
    a = 9.81m/s/s
    Vi = vsin23
    d = ? (assumed zero at landing)
    t = ?
    2. Relevant equations
    t = d/v
    at = vf - vi
    d = vi + 1/2at2

    3. The attempt at a solution

    t = 8.59m / vcos23

    from this point, I forgot explicitly the steps I took, but I subbed in what I knew for the initial velocities in an attempt to solve them. I ended up with:
    -0.3907v2 -0.3907v + 91.2 = 0

    I had included units in all my procedures, and the only units left in the equation were attached to the 91.2 and became m2s-2 (metres squared per seconds squared). I went on and used the quadratic formula anyway, despite the unit anomaly and came up with 14 m/s (rounded).

    I am not sure if this is even remotely correct, and if I am not even on the right track, could anyone show me how to properly solve it? It seems I'm lacking a bit in the math department, though I am at the top of my physics class (top three anyway).
  2. jcsd
  3. Nov 5, 2012 #2


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    Homework Helper

    I agree with your expression for t = from the horizontal d = vt equation.
    I don't know how you got the quadratic and I don't agree with its answer.
    For the vertical part of the motion, write Vf = Vi + at. Sub in the numbers you know. Sub in the expression for t from your horizontal equation. Use the trig identity that says
    2*sin(A)*cos(A) = sin(2A)
    and you should be able to solve for v without dealing with a quadratic equation.
  4. Nov 5, 2012 #3
    vi = -vf
    so, -vsin23 = vsin23 + -9.81m/s2 (8.59m / vcos23)

    Oh yeah, this is where I got the quadratic from...
    -v0.3709 = v0.3709 - 9.81m/s/s (8.59m/v0.9205)
    -v0.3709 - v0.3709 = -9.81m/s/s (v9.332m)
    -2(v)0.3709 = -91.5469m2/s2
    -0.7418v = -91.2vmm/ss (easier to type this way)
    -0.7418v = -91.2vmm/ss/-91.2mm/ss
    -0.7418v/-91.2mm/ss = v

    Now I'm confused again. If my math is right here, then the math on my test was wrong.
    And this is where my knowledge gap comes into play, given this equation:

    -v0.3709 = v0.3709 - 9.81m/s/s(8.59m/v0.9205)

    How does one solve for v?
    Last edited: Nov 5, 2012
  5. Nov 5, 2012 #4


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    Subtract .3709*v from both sides.
    Multiply both sides by v
    That gives you a v squared = calc.
    It seems to work out to 11.1 whereas I got 10.8 when I did not use the calculator until the very last step. Might be worth running it through with 4 digit accuracy.
  6. Nov 6, 2012 #5
    I asked my physics teacher about it (the question SHE put on the test) and we both couldn't figure it out. I just figured it out now: we were making careless math errors.

    -vsin23ms-1 = vsin23ms-1 + (-9.81m2s-2)(8.59m/vcos23ms-1)

    -vsin23ms-1 - vsin23ms-1 = (-9.81ms-2)(8.59m/vcos23ms-1)

    -2(vsin23ms-1) = -84.26m2s-2/vcos23ms-1

    (vms-1)-2vsin23ms-1 = -84.26m2s-2/vcos23ms-1 (vms-1)

    -0.7815v2m2s-2 = -84.36m2s-2/0.9205

    -0.7815v2m2s-2 = -91.54m2s-2

    -0.7815v2m2s-2 / -0.7815 = -91.54m2s-2 / -0.7815

    v2m2s-2 = 117.1m2s-2

    Then I just find the square root. The answer is 10.8ms-1 or 10.8m/s

    Thanks for the help guys, I probably would not have gotten the result with out the help of these forums!

    I was so close. Hopefully my teacher will give me part marks for my attempt on the test. I DID set up the equation right after all.

    Thanks again :D
  7. Nov 7, 2012 #6


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    It is a good habit to delay putting in the numbers until the very end, then doing one calculator step. It saves writing the numbers repeatedly and it means only one rounding off at the end. There are some exceptions where it is simpler to put some numbers in early.
  8. Nov 8, 2012 #7
    That's actually an excellent idea... my physics teacher did say that I should work on my critical thinking skills. Without sounding arrogant, I am at the top of my class but it seems that could use a bit of work. Thanks again for the help :D
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