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Projectile motion of a cannonball

  1. Oct 24, 2013 #1
    1. The problem statement, all variables and given/known data

    A cannon fires a cannonball 500 m down range when set at a 45 degree angle.At what velocity does the cannonball leave the cannon?

    2. Relevant equations

    Vyi=sin45 * Vi
    Vxi=cos45* Vi

    3. The attempt at a solution
    I don't know how to do this question?? Can someone help me?
    If u had time u could solve it but how to find that when u don't even know Vyi?

    I found this solution online.. I understand everything until line number 10. How did they get cos 45? Any help would be greatly appreciated

    Vyf = Vi sin 45+ at,

    at = - Vi sin45

    -gt = - Vi sin 45

    t = Vi sin 45/g

    And we defined t = 1/2 T or T = 2t

    T = 2*Vi sin 45/g

    Lets pull back up [eq1] and plug in this last equation into it.

    Vi cos 45 = R / T

    Vi cos 45 = R / (2*Vi sin 45/g)

    Vi^2 = g*R / (2*sin 45 *cos 45)

    Vi = sqrt (g*R / (2*sin 45 *cos 45))

    There you go, now just plug in your numbers.

    Vi = sqrt (9.81 m/s^2 *500 m / (2*sin 45 *cos 45))

    Vi = 70 m/s
     
    Last edited: Oct 24, 2013
  2. jcsd
  3. Oct 24, 2013 #2

    gneill

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    Staff: Mentor

    Cool12, please don't use text-speak here on PF.

    Write out your equations of motion for the projectile. What's special about sin and cos of 45°?
     
  4. Oct 24, 2013 #3
    Sorry, sin and Cos of 45 is the same
     
  5. Oct 24, 2013 #4

    gneill

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    Staff: Mentor

    Right. So what does that tell you about the velocity components?
    What's the sin or cos of 45°?

    Write out the equations of motion for the projectile.
     
  6. Oct 24, 2013 #5
    Sin /cos of 45 degrees is 0.707..
    But why are they multiplying sin45 by cos 45?
     
  7. Oct 24, 2013 #6

    gneill

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    Staff: Mentor

    That's one over root 2. It's a value worth remembering. 45° one of several common angles that it would be very helpful to remember the sin and cos for.

    ??? Who's "they"? The sin and cos of the launch angle are used to decompose the initial velocity into its separate vertical and horizontal components. Here the sin and cos are the same. So the component velocities are equal.

    Write the equations of motion for the projectile!
     
  8. Oct 24, 2013 #7
    thank you, I understand it now
     
    Last edited: Oct 24, 2013
  9. Oct 24, 2013 #8

    gneill

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    Staff: Mentor

    You should have mentioned that you'd edited your original post, adding additional information. I only looked back because I couldn't understand what you were referring to...

    Anyways, I looked at the derivation you posted, and while it works, it's a bit of an odd approach. It can be done quicker starting with the basic equations for projectile motion.

    But if you want to know what happened in that solution, I can help you there. You're worried about what happened around line 10, so let's take a look:
    At line 10, both sides were divided by cos(45), and both sides were multiplied by Vi. The g that divided the denominator on the RHS was moved to the numerator.
     
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