please explain why this trig solution is wrong

Discussion in 'General Math' started by kjm, Nov 15, 2003.

  1. kjm

    kjm Guest

    Hi all. I don't understand why my approach to this problem is flawed. Can
    someone explain it. Here is the problem and my attempted solution. (I know
    how to get the right answer but I don't know why a particular approach does
    not work)

    From Schaum's Beginning Physics I, Alvin Halpern
    Problem 3.43 A marksman is trying to hit a stationary target that is 100 m
    above the level of his rifle. He aims the rifle 5 deg above the horizontal,
    fires, and hits the target just as the bullet reaches the highest point in
    its trajectory. What is the muzzle velocity of the rifle? (not concerned)
    How far away, horizontally, was the target?

    The conondrum is in my particular solution to the 2nd of the two questions,
    how far away? I figured I could determine how far the target was by drawing
    a right triangle and applying trig like so:

    /| 100m
    / |
    5 deg

    tan 5 deg = 100 / x so x = 100 / tan 5 = 1143 m

    the correct answer is 2286 meters which is arrived at my applying well known
    physics formulae. The correct answer is 2286 m. This is given in the book
    and I confirmed it once using I forget what formulas.

    Illumination appreciated.

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    replace " AT " with "@"
    kjm, Nov 15, 2003
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  2. kjm

    Henry Guest

    The clue is in "highest point in its trajectory" suggesting that you
    need to take account of gravity
    Henry, Nov 15, 2003
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  3. kjm

    Tom Lorenzo Guest

    You have to understand more physics to get what the problem is about.
    Meaning you need to be less dependent on just finding a formula to apply and
    put some effort into that understanding. As Henry noted, the key is noting
    that the bullet strikes the target at the height of its trajectory. For
    these kinds of problems, you ignore air resistance to keep the calculations

    Think of it this way -- at the height of its trajectory, the vertical
    component of velocity of the bullet is zero. From that point of zero
    vertical velocity, the only thing affecting the vertical velocity of the
    bullet is gravity. So how long would it take a bullet dropped from 100
    meters to hit the ground? A further understanding of the problem is to
    realize that for level ground (no air resistance), the bullet will take as
    long to get from the ground to the height of its trajectory as it takes to
    fall from its highest point to the ground. By the way, your comment that you
    were not concerned about the muzzle velocity is a good indication you do not
    have a good grasp of the physics. Because after you compute how long it
    would take to fall 100 meters, the next step is to compute that muzzle
    velocity based on generating an equation for the vertical velocity of the
    bullet taking into account the "upward" velocity from the rifle and the
    "downward" velocity from the acceleration of gravity. Knowing the time it
    takes to fall 100 meters allows you to solve for the component of muzzle
    velocity in the vertical direction, and then the muzzle velocity, and then
    the component of muzzle velocity in the horizontal direction, and finally
    computing how far it traveled in the horizontal direction (because at that
    stage you know both the horizontal velocity and how much time it traveled at
    that velocity).

    - Sox
    Tom Lorenzo, Nov 16, 2003
  4. kjm

    kjm Guest

    I have received an email that explains how my approach is flawed. I did not
    take into account that the bullet travels along a parabolic trajectory when
    drawing my diagram. I am satisfied with this explanation but welcome more
    input if anyone wishes to say more.

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    kjm, Nov 17, 2003
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