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Projectile motion formula

  1. Jun 16, 2009 #1
    I would like to simply know what Δy=[(sin2θ)(Vi)]^2/2a represents..
    The formula only requires the angle and the initial velocity (along with the usual a=9.8m/s^2), so what height does the equation represent?
  2. jcsd
  3. Jun 16, 2009 #2
    The distance a projectile travels (over flat ground) is [sin(2θ)*(v^2)]/a, very similar to what you have there. Is there a chace you misread it?
  4. Jun 16, 2009 #3
    well that's the distance..
    I think I got it now though, it's possibly the maximum height over a flat ground.
  5. Jun 16, 2009 #4
    Oh yeah you're correct, i just googled it a bit and it came up. It is the maximum height you reach
  6. Jun 16, 2009 #5

    Doc Al

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    Staff: Mentor

    That's almost the formula for the maximum height of a trajectory. To correct it, replace sin2θ with sinθ.
  7. Jun 16, 2009 #6
    So one of the famous equations for the motion of a projectile is as follows:


    v = final velocity
    u = initial velocity
    a = g = -9.8m/s2
    s = distance travelled

    Now, for a projectile being fired at an angle, the vertical component of velocity is usin%, where % is the angle between the ground and the direction of projection.

    Rearrange your equation, with v=0 to get the maximum height attained by a projectile (note that in your equation, your v is my u).

    You get:

    s = u2/-2a = (usin%)2/-2a

    As a = -9.8, you can ignore the minus sign and you basically have your equation (except for the sin2% bit).

    Yes I don't know how to make greek alphabet symbols, so a % for an angle will do :D

    Hope that helps.
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