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Check if Solution is correct.

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Derive an equation for the horizontal range of a projectile with a landing point at a different altitude from its launch point. Write the equation in terms of the initial velocity, the acceleration due to gravity, the launch angle, and the vertical component of the displacement. Please check to see if it is correct!
    I tried to make the formulas as realistic as possible.

    p.s. the "/" tend's to mean a fraction. a/b = a/b

    2. Relevant equations

    x = vcosΘt, t = v/cosΘ (1)

    y = vsinΘt - 1/2 gt2 (2)

    3. The attempt at a solution

    Solving (1) for t and substituting this expression in (2) gives:

    y = xtanΘ - (gx2/2v2cos2Θ)

    y = xtanΘ - (gx2sec2Θ/2v2) .....(Trig. identity)

    y = xtanΘ - (gx2/2v2) * (1+tan2Θ) .....(Trig. identity)

    0 = (-gx2/2v2 * tan2Θ) + xtanΘ - (gx2/2v2) - y .....(Algebra)

    Let p = tanΘ​

    0 = (-gx2/2v2 * p2) + xp - (gx2/2v2) - y .....(Substitution)

    p = -x ± √{ x2 - 4(-gx2/2v2)(-gx2/2v2 -y)} .....(Quadratic formula)
    ,,,,,,,,,,,,,,,,,,,2 (-gx2/2v2)

    p = v2 ± √{ v4 - g(gx2 + 2yv2)} .....(Algebra)
    ,,,,,,,,,,,,,,,,,,,,,,,,,gx

    tanΘ = v2 ± √{ v4 - g(gx2 + 2yv2)} .....(Substitution)
    ,,,,,,,,,,,,,,,,,,,,,,,,,gx

    Θ = tan-1 [v2 ± √{ v4 - g(gx2 + 2yv2)}]
    ,,,,,,,,,,,,,,,,,,,,,,,,,gx
     
  2. jcsd
  3. Oct 2, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    Your algebra all looks good! But it seems to me you are asked to find x as a function of the initial speed, angle and y. That would mean starting with x = vt*cos(theta) and trying to solve the y equation for t in order to eliminate t in the expression for x.
     
  4. Oct 4, 2009 #3
    Hmm. I see, I guess I just started solving for anything else.
    Do you have any suggestions how I can do this? or how it looks like?
     
  5. Oct 4, 2009 #4
    Let the desired horizontal range be h while the corresponding vertical height is k. We have:

    [tex]h= v_{0}\cos(\theta)t \iff t = \frac{h}{v_{0}\cos(\theta)} \, * [/tex]

    And

    [tex]k = v_{0}\sin(\theta)t + \frac{1}{2} g t^{2}[/tex]

    Use the substitution given by (*) to get:

    [tex] k = v_{0}\sin(\theta) \left( \frac{h}{v_{0}\cos(\theta)} \right) + \frac{1}{2} g \left( \frac{h}{v_{0}\cos(\theta)} \right)^{2} [/tex]

    [tex]k = \tan(\theta)h + \frac{gh^{2}\sec^{2}(\theta)}{2v^{2}_{0}}[/tex]

    This is as far as you need to go since you have a relationship in terms of all the variables asked for in the question. It might be possible to isolate h in the equation, though I have no idea how it would be done.
     
  6. Oct 4, 2009 #5
    Whew.. all that work for nothing.
    Thanks mate!
     
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