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What's the proof

  1. Mar 8, 2005 #1
    I know that a projectile flyes the farthest, when it is louncht under a 45 degree angel. But I don't know how to proofe this. Any help would be apreeciated.
  2. jcsd
  3. Mar 8, 2005 #2


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    Well,then can u write down the equations of motion...?

  4. Mar 8, 2005 #3


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    45 degrees assuming no air resistance!

    If the angle is θ degrees, initial speed v, then the horizontal and vertical components of initial velocity are vcos(θ) and vsin(&theta) respectively.

    acceleration is 0 horizontally, - g vertically so
    velocity (t)= vcos(θ) horizontally, vsin(θ)- gt vertically.

    position(t)= vcos(θ)t horizontally, vsin(θ)t- (1/2)gt2 vertically. (taking (0,0) as starting point.)

    1. Solve for the time of impact (i.e. set height= 0, find non-zero t solution- it will depend upon θ).

    2. Find horizontal distance at that time (again, it will depend upon θ).

    3. Find the value of θ that maximizes that (differentiate with respect to θ, set equal to 0).
  5. Mar 8, 2005 #4
    Thenks! I got to the 2'nd step myself but the third one is realy helpful. I didn't think about useing differentilas, becouse I only started takeing them a few weaks ago and I still don't find my way around very well. Thenks agein.
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