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Angle when a projectile is launched at a height from the ground

  1. Jun 12, 2013 #1
    This a question from projectile motion. This is NOT "homework"; it's a concept that got me thinking.

    When a projectile is launched from the ground level and it falls back on the same (ground) level, the horizontal displacement is maximised if the angle of projection (with respect to the horizontal) is 45 degrees.

    BUT when the projectile is launched from a height above the ground, and it falls onto the ground level, the maximum horizontal displacement is covered if the angle of projection is slightly less than 45 degrees.


    My question is, what relation can we derive to determine the angle theta (w.r.t. the horizontal) at which the projectile should be launched from that height h so as to get maximum horizontal displacement M? That is, what is the relation between theta, h, R and M as per the illustration (attached)?
    R is the horizontal displacement if the angle of projection at h were 45 degrees.

    Attached Files:

  2. jcsd
  3. Jun 12, 2013 #2


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    The basic equations you need to work with are (ignoring air resistance):
    0 = h + vtsinθ - gt2/2 (process stops when you hit the ground)
    d = vtcosθ

    d = horizontal distance, t = time of flight, v = initial speed of projectile, g = accceleration due to gravity

    Eliminate t to get d as a function of θ. Use calculus to maximize d.
  4. Jun 12, 2013 #3
    Try this:
    ##sin \alpha = \frac{1}{\sqrt{2 - \frac{2gh}{u^2}}}##

    (Negative value of 'h' needs to be applied for points below. )

    (I know I am not supposed to post answers here but I guess it is easier than telling you how to work it out.)

    Deriving that is your homework.
    Last edited: Jun 12, 2013
  5. Jun 12, 2013 #4


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    Gold Member

    But it certainly could be, so to the homework forum with it.
  6. Jun 14, 2013 #5
    Thanks. That does give me the relation I wanted!
    (Btw, I got this from Wolfram|Alpha.)
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