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Homework Help: Maxiumum height and range

  1. Aug 20, 2007 #1
    A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection?

    If I let A = maximum height (or h), than horizontal range (or R) = 3A.

    I suppose I can also substitute 1 for h and 3 for R. There is some relationship between the horizontal velocity and the initial vertical velocity that allows for this feat to be accomplished. The angle must be of about medium amount, about 40-60 degrees?

    Equations include the components of vf = vi + at, components of rf = ri + vit + .5a(t^2), equations for range and height:

    h = ((vi^2)((sinθ)^2))/(2g), where g = -9.8 m/s^2
    time it takes for particle to reach max height: th = (vi(sinθ))/g
    R = vxi2th = ((vi^2)(sin2θ))/g

    θ = angle of projection, but all of these equations require some knowledge of the initial velocity, which I don't have. And the equations involving that include time, which I also do not have. I think I'm not getting something important from the information already given. The point at the beginning is (0,0), while at maximum height, the coordinates are (1.5,1) or (3A, A).

    So, now I think that I can use the rules of parabolic equations to figure this out. But, I don't think they want me to figure this problem out like that.
  2. jcsd
  3. Aug 20, 2007 #2


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    What you need to find are the ratio of horizontal and vertical velocities at the start.
    For the horizontal component there is no accelaration so s(h) = v(h) * t
    Vertically you have the famous s = ut + 1/2 at^2 use this for the vertical motion. Remember the time to reach the top of the parabola is half the time to complete the flight.
    You should be able to end up with values for v(horiz) and v(vert) then a simple bit of trig will give the angle.
  4. Aug 20, 2007 #3
    But if I do not know the time, nor the value of the initial velocity vectors, how can I figure it out?
  5. Aug 20, 2007 #4


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    You are going to get equations for the initial vertical and horizontal velocity.
    They might have 't' in them but you are going to take a ratio of the two velocity components to get the overall direction so these will cancel out.

    Or you can use energy conservtion:
    Vertically at the start it has only ke = 1/2 m v(v)^2
    At the top it has only potential pe = m g h
    Mass cancels so;
    g R = 0.5 * v(v)^2 or V(v) = sqrt(2 g R )

    You can then use s = ut + 1/2 at^2 to get the time.
    Then use this time to get the horizontal speed, remember that 't' is the time to get to the top of the arc so is half the time to travel the horizontal distance.
  6. Aug 21, 2007 #5
    I am still having trouble with this problem, although I sincerely appreciate your help.

    For x: 1.5 = vxit or t = 1.5/vxi

    As for y: 1 = vyit - 4.9(t^2)

    I understand that while x and y are of independent paths, they both have something in common, usually either a time or an angle. In this case, it is time. I do not think I have to alter the variable of time in any way, as the final x and y points I am dealing with are the ones at the pinnacle of the parabola. Therefore, the time should only be of half the trajectory. Correct me if I am wrong in following this.

    I've converted the x part into t = 1.5/vxi, and tried plugging it into what I had for y, and I don't think that's right either and, well, yea.

  7. Aug 21, 2007 #6


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    You've already done most of the work in your first post. You've calculated the height in terms of v and [tex]\theta[/tex]. You've calculated the range in terms of v and [tex]\theta[/tex]

    divide the h equation by the R equation... v^2 cancels out. h cancels out since R = 3h. Now you can get [tex]\theta[/tex]
  8. Aug 21, 2007 #7
    Ooooookay, so:

    1/3 = (sinθ)^2/(2sin2θ)


    (3/2) = sin2θ/(sinθ)^2

    But, that requires sin2θ to equal three, and that is not possible.
  9. Aug 21, 2007 #8


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    expand sin2θ then simplify.
  10. Aug 21, 2007 #9
    OH crudmuffins!

    I completely forgot about...well, the 23 basic trig identities I had (note the past tense) memorized about a year ago.

    All right, all right. It's about 37 degrees. Thank you...so much for you patience and help. :D I appreciate it greatly.
  11. Aug 21, 2007 #10


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    It's not 37 degrees, you're close through... you made a small mistake. you mixed up 3/4 and 4/3... :wink:

    The answer's 53 degrees.
    Last edited: Aug 21, 2007
  12. Aug 21, 2007 #11


    Well, this is embarrassing.

  13. Aug 21, 2007 #12


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    Don't be embarrased. I've lost count of how many times that type of thing has happened to me! :smile:
  14. Aug 21, 2007 #13
    M'kay. I actually got that answer you did.

    I took the tangent instead of the cotangent, the way I divided over all the numbers.
  15. Aug 21, 2007 #14


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    Cool. :cool:
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