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Optimal Trajectory for a ball thrown WITH INITIAL HEIGHT

  1. Jan 30, 2009 #1
    A ball being thrown, ignoring drag or wind, has an initial height, initial velocity, and experiences the acceleration due to gravity.

    When a ball is launched from the ground, y_0 = zero, complementary angles (for example 20 and 70 degrees) will produce the same distance for the same initial velocity, and the maximum distance is acheived at 45 degrees.

    However, with an initial height an angle of more than 45 degrees will result in a farther distance.

    My question then is what is the relationship between angle of trajectory and initial height to give the farthest distance for a constant velocity. It seems like there must be one... i've been drawing some graphs to try to figure it out. The time for the ball to reach maximum height will stay the same for every angle, and i think total travel time (counting the negative time) will be the same as well...

    Pretty curious, let me know what you think.
  2. jcsd
  3. Jan 30, 2009 #2


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    The opposite is actually true. When the vertical displacement is negative, the optimal angle is less than 45 degrees, whereas when it is positive, the optimal angle will be more than 45 degrees.

    I don't think drawing graphs is going to get you a solution to the problem, so here's a helpful way to think about it:
    You want to maximize the distance function, which can be written as [tex]x(t,\theta)=v_{0}cos\theta t[/tex].
  4. Jan 30, 2009 #3

    yes that's what i meant, a positive initial height will give an optimal angle of more than 45 degrees, and depending on how high its released from will depend on the angle?

    for 1 velocity of which the x and y component are related and dependent on this mystery angle I want to figure out.

    the cosine function has min at 90 and a maximum at 0 degrees, but i know just looking that the optimal angle is above 45 degrees. i can't tell how to reconcile this angle value with the differing values for t caused by a different total path length when fired from a different height.

    i was trying to maximize it by taking the equation for y, [tex]y(t,\theta)=y_{0} + v_{0}sin\theta t + 1/2at^2[/tex]

    solving for the angle that gives a maximum t and putting that in the distance, but of course that would be a straight up 90 degree thrown and it wouldnt go anywhere.

    maybe, if i solve for t in the above equation, and put that into the distance formula, we'll have distance traveled in terms of initial velocity, constant; acceleration - constant; and then just the angle and the height, whose whole relationship i'm trying to get at. looking at this equation though, and even choosing a value for height, its not immediately clear how to find the maximum angle. we know have x in terms of the angle at least ,


    which is what i wanted to know, so maybe set the derivative of the function and set to 0 to find the theta which gives the function a maximum...

    hmmm that sounds pretty good.
    Last edited: Jan 30, 2009
  5. Jan 31, 2009 #4


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    Your language is confusing me but I think you have the idea here, you should just go through the process:

    [tex]y(t,\theta)=h_{0}+v_{0}sin\theta t - \frac{1}{2}at^{2}[/tex]
    [tex]x(t,\theta)=v_{0}cos\theta t [/tex]

    Am I right in thinking this is what you are thinking of? The relationship y/x should probably aid in getting an explicit function of x as a function of theta.
  6. Jan 31, 2009 #5

    y(t,\theta)=h_{0}+v_{0}sin\theta t - \frac{1}{2}at^{2}

    when the object hits the ground, y = 0. solving for t in this equation then with the quadratic equation, and placing in place of t here:

    x(t,\theta)=v_{0}cos\theta t

    gives x in terms of theta

    x(\theta)= v_{0}^{2}sin\theta cos\theta /{g} +- \sqrt{(v_{0}^{2}sin^{2}\theta - 2 g h)} / {g}{

    with v_{0} and g constant...
  7. Jan 31, 2009 #6
    FWIW, when air resistance is INCLUDED, the optimal angle of initial trajectory is 41 deg.
  8. Jan 31, 2009 #7
    I don't want to bore anyone with the practical, but if you are firing an arrow and you are above the target you aim below it and if you are firing up you aim above it, interesting example of targeting with a non stationary projectile. It all depends on the effect you want, for example a cannon against castle walls would essentially be your stereotypical maths from above, but a howitzer you might want to target much more accurately you would fire entirely differently. A mortar or a trebuchet of course is fired still differently and is better able to attack targets which are behind walls or above the field of fire or indeed close to. The guns of WWII battleships could hit a m^2 target from 5-10 miles off shore. Now that's maths in action. :smile:
  9. Jan 31, 2009 #8
    Following Arod's method to get x(theta), one would find the maximum by differentiating x with respect to theta and solving for the angle at which this derivative is zero. The resulting expression must be used carefully to obtain a sensible result because of intermediate steps involving square roots and inverse trigonometry equations:

    [tex]\theta =\cos ^{-1}\left(\frac{\sqrt{2 g h+v^2}}{\sqrt{2 g h+2 v^2}}\right)[/tex]

    Note that this reduces to pi/4 when h = 0.
  10. Jan 31, 2009 #9
    so beautiful and so serene!
  11. Feb 10, 2009 #10
    How can you say that if you haven't specified the shape/mass of the projectile? Without air resistance, we just "throw" points, but drag is dependent on a few characteristics of the object. You can't include drag unless you define these characteristics.

    Actually, wait. Maybe your right. Just because those characteristics change the maximum distance of the projectile, doesn't mean it changes the optimal angle, necessarily.

    So, when deriving the optimal angle to launch a projectile in a system with drag, does the object's mass, etc. drop out of the equation?
    Last edited: Feb 10, 2009
  12. Feb 11, 2009 #11
    Here is an excellent article on optimal trajectory for throwing a soccer ball. It pays attention to the limitations of the body's kinematics. In this case, the optimal angle is only 30 deg from horizontal.

  13. Feb 11, 2009 #12

    Do not ever make this mistake. When the angle of trajectory is changed, you are changing how much energy is being put into each dimension. Remember that when doing mulitdimensional problems, the x, y, and z dimensions are all independent of one another. When the angle of trajectory is greater than 45 degrees, there will be more energy in the y direction that translates into motion. When launched at an angle less than 45 degrees, there will be more energy in the x direction. This is only true when the displacement is zero.
  14. Feb 12, 2009 #13
    This is actually an extremely complex problem that is usually made simple by discounting any effects that could potentially change the outcome of the trajectory. Shame in real life when you're trying to lob a blazing ball of pitch into the church roof of an enemy castle, you can't write a note to the butcher apologising for a sudden shift in a vector quantity you hadn't quite accounted for. Or apologise to the president of Nicaragua for missing him due to a sudden rain shower that deviated the ballistics trajectory and made you hit his daughter.
  15. Jan 11, 2012 #14

    How did you solve that algebraically? I've tried simplifying for ages, but seems I can't find the way to get an easy expression for theta.

    I'm certain that my derivative is correct but I can't get rid with all the squareroots when simplifying.:S
    Last edited: Jan 11, 2012
  16. Jan 13, 2012 #15
    Hey guys, I looked at this thread out of curiousity and it really set me thinking.

    I don't see why an angle bigger than 45 degrees would make your distance farther. As far as l can tell the parabola will just be horizontally displaced by h0, which will make the trajectories identical until it reaches h0 above the ground where from it's obvious that the 45degree shot will go farthest. What am I missing out on?

    Are we considering the fact that the object itself might be raised a little by making the angle bigger? I mean if we have a cannon of length L then obviously the cannonball will be at a higher and higher positions as our angle goes to 90degrees.
  17. Jan 13, 2012 #16
    Nvm, I figured it out. Pretty beautiful result from such ugly equations..:)

    Not sure if I understood your first question right, but:
    Making the angle bigger than 45 degrees is only benefiting in range if the target area is higher up than the objects starting position. This is because you want to have a maximum combination of the time the object is in the air and the velocity in your x-direction, this due to the fact that the gravitation have a force that accelerates the object down, and "too" long time in the air will just be a loss if you dont have enough speed in x-direction.

    No we are not considering the fact that the object itself might gain or loose some height when we change the angle. When you consider a cannon the height difference will probably be rather small(depending on how big the cannon is), and ofc, if you want a better answer you will have to take this into your calculations.

    hope this makes sense. xD

    Many interesting things happen when you involve the dragforce from air because then the optimum angle will depend on so much more than speed and height. For example if you have a spinrate on the object it will be affected by aerodynamical forces, if you have a backspin this force will act as a lift force on the object, hence the object will stay in the air for a longer time, this means that lowering the angle will benefit the range of the trajectory.
    Last edited: Jan 13, 2012
  18. Jan 13, 2012 #17
    Who said something about target area being higher? Aren't we discussuing the situation drawn in the attached file? If not please tell me what I miss out on - or even better: Correct my picture :)

    Attached Files:

  19. Jan 13, 2012 #18
    The formula takes on both scenarios, if the situation is the one in your picture the "h" is positive and if its the opposite the "h" is negative. This implies that if ur target area is below the starting position, the bigger height difference, the smaller angles for optimating trajectory. And with the target area being above starting position, its the opposite.

    Great pic;) Btw I might be wrong, actually it says "Optimal Trajectory for a ball thrown WITH INITIAL HEIGHT", this includes "negative" initial height?
  20. Jan 13, 2012 #19
    ahh I get it now. When the height gets big enough in comparison to the the maximum height of the trajectory, it too gets something to say about how long time the ball stays in the air...
    But on my picture the optimum angle is below 45 right? else I misunderstood something again.
  21. Jan 13, 2012 #20
    Yeah you are right, no misunderstanding there. ;D
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