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Projectile Motion: Formula for range for a given angle

  1. Aug 2, 2013 #1
    I was watching a YouTube video on projectile motion:

    It says that the range is given by the formula:

    d = v*cosθ*t

    It also says that the time to reach the maximum height is given by the formula:

    t = [itex]\frac{v*sinθ}{a}[/itex] where v*sinθ is the initial vertical velocity

    Well, not necessarily, but I derived that from:

    v = u+at

    So you multiply that by 2 to get the total time which gives you:

    t = [itex]\frac{2*v*sinθ}{a}[/itex]

    Put that into the original formula and you get:

    d = 2*v*cosθ*([itex]\frac{v*sinθ}{9.81}[/itex])

    d = [itex]\frac{2*sinθ*cosθ*v2}{9.81}[/itex] -- meant to be 2*sinθ*cosθ*v2/9.81

    My aim was to plot the range of the projectile at a given angle. However, when I plot this graph using Graphmatica, it is just an oscilating graph, resembling the sine or cosine functions...

    Where have I gone wrong? Either a simple mistake, or this was never going to work in the first place.

    Thanks for any help.
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Aug 2, 2013 #2
    As far as I understand you have done every thing correctly. As for the graph it is varying sinusoidally as you haven't specified the limits of the angle; when you throw something the angle will be between 0 to 90 degrees the graph should accurately give the distance in that interval.
    An interesting point is as the equation can be reduced in just terms of sine the same range can be achieved for two different angles in 0--90° interval...except For 45° when the range is max.
  4. Aug 2, 2013 #3
    Thanks for the reply. I have set the domain of the function from 0-90, however it still doesn't seem to be the right shape. It is still the oscillation, oscillating about 30 times between x = 0,90.
  5. Aug 2, 2013 #4
    Greetings again,
    The math is flawless, I entered the equation in an android grapher it checked out.
    Are you sure the scale of x axis is in degrees and not radians for reference 90 degree equals 1.58 approximately and 45 around 0.785 .....if thats not the problem I really have no idea whats wrong.
  6. Aug 4, 2013 #5
    Ok, it turns out it was in radians. I have now converted it to degrees, and with the help of a friend, is now in a more refined format:

    d = sin([itex]\frac{∏*θ}{90}[/itex])*vx2/9.81

    and for Graphmatica: y = sin(pi/90*x)*10^2/9.81 {x: 0,90}

    Thankyou very much for you help.
    Last edited: Aug 4, 2013
  7. Aug 4, 2013 #6
    Welcome :)
  8. Aug 4, 2013 #7


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    Gold Member

    Fun fact: In real life, around 1900 Krupp artillery testers discovered by accident that gun elevations above 45 degrees improved the range considerably. It turns out that air resistance falls off with increased altitude, so a steeper angle gets the projectile to a low friction zone faster.
    In a vacuum of course the 45 degree solution is the right one.
  9. Aug 5, 2013 #8
    Oh, because higher altitudes have lower air pressures, but that would only really matter if the projectile goes to considerable heights, and it probably wouldn't be much past 45° unless it was travelling really fast.
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