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Projectile Force and Velocity?

  1. Jun 1, 2015 #1
    • Missing template due to originally being posted in different forum
    Hi guys, Please, someone can help me to solve some simple equations?
    I Need to shoot a projectile with 20 grams at 15 meters and 15°.
    So, I'm triyng to calculate the velocity and Force and what I will need to make this succesfull.

    My Calcs:

    *************************************************************************
    Velocity= √(R*g)/sen θ
    V= √(15m * 9,81) /sen15
    V= 46,86 m/s
    ***
    I= ΔQ or Qend - Qinitial. (Q = moviment quantity. Q= m.V)

    I= (0,020kg * 46,86 m/s) - (0,020kg * 0 m/s)
    I= 0,937 N.s

    ***
    (I = Impulse // I= Force * Time)

    0,937 N.s = Force * 0.1s
    Force = 0,937 / 0,1 = 9,37 N.
    **************************************************************************

    This means I need to put just 9,37 Newtons on 0,1 seconds to shoot a 20 grams projectile at 15 meters and 46,86m/s? (at 15°)

    Thank you for your attention and sorry for my bad english.
     
  2. jcsd
  3. Jun 1, 2015 #2
    Isn't the range formula with a sin(2θ)?
     
  4. Jun 1, 2015 #3

    BvU

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    Hello AA, welcome to PF :smile: !

    You want to be a bit more clear in your mission statement:
    1. is the 15 degrees wrt horizontal ?
    2. Is the 15 m in a horizontal direction ?

    Your first calc equates potential energy from height with kinetic energy: ##{1\over 2} m (v\sin\theta)^2 = mgh ## hence ## v = \sqrt{2 g h}/\sin\theta ##.
    So like Nasu I think there is a factor 2 missing, only in a different place.
    Indeed your 46.86 m/s will only get the object to a height of 7.5 m (that factor 2).
    But by then it has already travelled 56 m and it hits the ground at 112 m. Is that the idea ?
     
  5. Jun 1, 2015 #4
    I looked my notes but didn't has any "2" in the formula, my error.

    The intention is throw the projectile 15m in a horizontal direction like that: http://imgur.com/rswHEH0 .

    About the formula : √2gh / sin θ.
    This will not just increase my velocity?
    About the 15°, how can I calculate with 0° (parallel to the ground) ?

    Thank you for the help Natsu and BvU!
    Have a good day guys...
     
  6. Jun 1, 2015 #5
    If it's about horizontal range, then you can look up the right formula here, for example: http://en.wikipedia.org/wiki/Range_of_a_projectile
    If you launch from ground level, the second formula will apply.

    If you want to have 0 degrees angle you need to launch from some height, don't you?
     
  7. Jun 2, 2015 #6
    I see now, thank you @nasu and @BvU .
    I will do the calculations.!
     
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