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Homework Help: Calculate the biggest horizontal distance that can be reached

  1. Aug 15, 2011 #1
    An object is thrown of a given H height by a given Vo initial velocity. Calculate the biggest horizontal distance the object can reach from the starting point.

    It was a questioon from the national physics olympiad in my country.

    I tried to solve, but I was not able to

    if the object is thrown by a [itex]/alpha[/itex] angle, we can easy calculate

    Horizontal Velocity [itex]Vx = Vo cos(\alpha)[/itex]
    Vertical Velocity [itex]Vy = Vo sin(\alpha)[/itex]

    The height in function of time

    [itex]h=H+Vo sin(\alpha) t - (1/2)gt²[/itex]

    If h = 0, Solving we get

    [itex]t = \frac{Vo sin(\alpha) + \sqrt{ Vo²sin²(\alpha) + 2gH} )}{g} [/itex]

    So horizontal distance A =

    [itex]A = Vo cos(\alpha) (\frac{Vo sin(\alpha) + \sqrt{ Vo²sin²(\alpha) + 2gH} )}{g} )[/itex]

    Derivating this and equaling to 0 we get the \alpha for max velocity, but I've tried this and I got a 6th power equation, even Wolfram could not solve it

    Do you think there's a easier way or the problem should be canceled ?
  2. jcsd
  3. Aug 15, 2011 #2


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    It sounds like you did the reasonable things after differentiation, such as multiplying through once by the radical to eliminate it from two of the terms, moving those terms to the other side of the equation, and squaring again to completely eliminate radicals. [You can also remove a factor of Vo/g .] A remaining hope for simplification may come from using trigonometric identities.

    Would you mind writing out your equation, so we can see if we're talking about the same result? There is no reason the result for the angle producing maximum range should be "pretty", since the trajectory is asymmetrical.
  4. Aug 15, 2011 #3
    I've got

    [itex]\frac{Vo² Cos(x)² }{g} - \frac{Vo² Sin(x)²}{g} + \frac{Vo³ Cos(x)² Sin(x)²}{g \sqrt{2 g H + Vo² Sin(x)²}} - \frac{Vo Sin(x) \sqrt {2 g H + Vo² Sin(x)²}}{g}[/itex]

    derivating that

    I don't see an easy equation to solve, even to replace the x value in the A then

    Do you have an alternate formule? Because I don't rhink this problem is solveable for an olimpyad.

  5. Aug 15, 2011 #4

    Ray Vickson

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    Substituting s = sin(alpha) and cos^2(alpha)=1-s^2 in the expression for dA/d(alpha) and equating the derivative to zero gives an equation involving s, s^2, s^3 as well as a square root of a quadratic polynomial in s. Maple 11 easily gets the solution

    s = V0/sqrt(2V0^2 + 2gH).

    This can be verified, by substituting it back into the equation and showing (using Maple) that we get zero; also, we can plug in some numerical values for V0, H (and set g to its known value) and plot the numerator of dA/d(alpha) as a function of s (0 <= s <= 1), so can verify that the Maple solution is correct and unique.

    This problem seems too hard for an olympiad in which powerful computer algebra programs are not available.

  6. Aug 15, 2011 #5


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    I disagree at only one place: I think the term with the radical in the denominator only has one power of sine in the numerator. Removing a factor of Vo/g leaves

    [tex]V_0 cos^{2}\alpha - V_0 sin^{2}\alpha - sin\alpha \sqrt{2gH + V_0^{2} sin^{2}\alpha} + \frac{V_0^{2} cos^{2}\alpha sin\alpha}{\sqrt{2gH + V_0^{2} sin^{2}\alpha}} = 0 . [/tex]

    The use of the Pythagorean Identity suggested by Mr. Vickson is a good idea. If we also multiply through by the radical, we have

    [tex]V_0[1 - 2 sin^{2}\alpha ] \sqrt{2gH + V_0^{2} sin^{2}\alpha} = sin\alpha [2gH + V_0^{2} sin^{2}\alpha] - V_0^{2} (sin\alpha - sin^{3}\alpha ) . [/tex]

    We can make a bit of simplification before squaring both sides again:

    [tex]V_0[1 - 2 sin^{2}\alpha ] \sqrt{2gH + V_0^{2} sin^{2}\alpha} = [ 2gH - V_0^{2} ] sin\alpha + 2 V_0^{2} sin^{3}\alpha . [/tex]

    I can continue working on this, but I'm not seeing a clean result just yet. This might be too hard for an olympiad, but made another idea will arise...

    EDIT-- Yes, try it from here: after squaring both sides, you will have a polynomial in powers of [tex] sin^{2}\alpha [/tex] and the term containing [tex] sin^{6}\alpha [/tex] cancels! It looks like you have only a quadratic equation to solve after substituting [tex] x = sin^{2}\alpha . [/tex]
    Last edited: Aug 15, 2011
  7. Aug 16, 2011 #6
    Thanks dynamicsolo and Vickson, I did that and got the same result as Mr. Vickson

    [itex] \frac{Vo}{ \sqrt{2Vo² + 2gH} } [/itex]

    But as Mr. Vickson said, it's very difficult even to find the result with computers, I would understand if this was a problem of IPhO trainning as I did the las year, where he gives 3 problems to you and 5 hours to solve. But it's not, It's a problem from the second phase, Brazilian Olimpyad is divided like that:

    -1st phase, 20 tests (you had to score 9 this year), it's PRETTY easy, almos t all decent student does

    -2nd phase, 4 tests of direct answer (only the answer counts) and 4 of dissertative anwer (the resolution counts), it's a little bit difficult, from the initial people, only 3% get classified

    -3rd phase, 8 tests of dissertative answer (60% of the grade) and a experimental problem (40% of the grade), a little bit difficult than the last one, 50 people of each grade get cassified, where the first 20 do IPhO training

    I mean, do you think they applied correctly this question in the test? Or do you think I should email them asking for the cancelement?
  8. Aug 16, 2011 #7

    Ray Vickson

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    The question is relatively easy if we regard it as an equality-constrained optimization problem, and solve it using a Lagrange multiplier method. (That would be OK on an olympiad, or even as a homework problem.) Using the notation v = V0, c = cos(alpha) and noting that sin(alpha) = sqrt(1-c^2), the problem is:
    maximize vct, subject to H + vt*sqrt(1-c^2)- gt^2/2 = 0 (and 0 <= c <= 1). The Lagrangian for this optimization (NOT the Lagrangian in the sense of Mechanics!) is L = vct + u*[H + vt*sqrt(1-c^2) - gt^2/2], where u is the Lagrange multiplier. The optimality conditions are dL/dc = 0 and dL/dt = 0. Now dL/dc = vt*[1 -uc/sqrt(1-c^2)], so setting this to zero gives a simple equation for c, whose positive root is c = 1/sqrt(1 + u^2). Substitute this into dL/dt and simplify, to get dL/dt = [v + v*u^2 - ugt*sqrt(1+u^2)]/sqrt(1+u^2), so equating this to zero gives a simple solution for t: t = v*sqrt(1+u^2)/(ug). Putting these values of c and t into the constraint, we get the equation (v^2 + 2gH)*u^2 - v^2 = 0. We must choose the positive root (to make t positive), which is u = v/sqrt(v^2 + 2gH). Now put this value into t and c, and substitute into Xmax = vct, to get Xmax = v*sqrt(v^2 + 2gH)/g.

    Note that we never had to solve any equations more complicated than variable^2 = number.

  9. Aug 16, 2011 #8
    Sorry ms Vickson but we don't learn Lagrange in high school

    For me I think the problem is too difficult for 2nd and 3rd high school years, I've already emailed them to see if they will cancel
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