1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    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!

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

    User Avatar

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook