1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding Max Range With 2 Angles

  1. Oct 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Today my teacher asked me "What are the best 2 angles to use when wanting to get maximum horizontal distance while on an elevated surface of 50m."

    As in Dx. What would these 2 angles have to be for the projectile to go the furthest distance horizontally and what would the angles have to be for the projectile to have the highest vertical height.

    How could I figure out what the 2 angles are? Is there a certain formula you would use and just plug it in?

    What would X and Y have to be.


    2. Relevant equations

    No clue since this wasn't homework and my teacher just asked me if i could try and figure out how to solve it.
    dR/d∅ = V2/g [cos∅ (cos∅ + (sin∅ * cos∅) / √sin∅2 + C ) - sin∅(sin∅ + √sin∅ + C))

    3. The attempt at a solution

    I tried finding different equations, but this is the only one I could find and it didn't work.
    I know one angle can't be above 45 since both angles have to total to 90.
    Last edited: Oct 24, 2013
  2. jcsd
  3. Oct 24, 2013 #2


    User Avatar
    Gold Member

    If a projectile is fired at an angle θ to the horizintal with velocity v, then the vertical component is v sin(θ) and the horizontal is v cos(θ). The distance travelled is always v T cos(θ) where T is time spent in the air. The time spent in the air can be found like this. The vertical velocity is

    V(t) = v sin(θ) -g t

    and this is zero when t = v sin(θ) / g which is the at the highest point. The vertical distance is

    D = H + v t sin(θ) - (1/2)g t2 where H is the initial heght. So at the highest point

    D = H + v sin(θ) v sin(θ) / g - (1/2)g ( v sin(θ) / g )2 = H + (1/2)v2 sin(θ)2/g.

    The time in the air is

    T2 = 2D/g = (gH + v2 sin(θ)2)/g2

    and the square of the horizontal distance is v2cos(θ)2[(gH + v2 sin(θ)2)/g2]

    With H = 0 it is obvious that θ = 45o maximises the distance. But otherwise there's no easy answer.

    I found the general answer by differentiating the horizontal distance and finding the value of θ that zeros it.

    cos(2θ) = g H / v2

    and when H=0, the maximum is θ = 45o which agrees.

    The angle at landing α is given by tan(α) = (vertical velocity)/v cos(θ)= √(gH + v2 sin(θ)2))/(vcos(θ))

    and again if H = 0 the angles are equal.

    [Edit : corrected an error in the last expression]
    Last edited: Oct 25, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted