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Projectile problem

  1. Feb 21, 2005 #1

    tony873004

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    *** edit to fix a few mistakes pointed out by Davorak in the 2nd post to this thread ***

    A cannon is used to launch capsules of fire retardant chemical onto a fire. The cannon is 1.0 km from a cliff that is 0.5 km high. The fire is 1.0 km from the base of the cliff. If the capsules are launched at an angle of 45 degrees, what initial speed v is necessary?

    List what I know:
    The cannon is 500 meters above the height of the fire.
    The cannon is 2000 meters from the fire in the x direction.

    [tex]v_{iy}=sin45*v_{i} [/tex]
    [tex]v_{ix}=cos45*v_{i} [/tex]

    Now try to solve it:

    [tex]t_{x} = t_{y}[/tex]

    [tex]t_{x} = \frac{2000}{v_{ix}}[/tex]

    Using the formula:
    [tex]y_{f} = {y_{i}+v_{iy}t-\frac{1}{2}gt^2=0 [/tex]

    Construct a quadratic equation:
    [tex]t_{y} = \frac{-v_{iy} \pm \sqrt{v_{iy}^2-4*4.9*yi}}{2(-\frac{1}{2}g)}[/tex]

    Set it equal to [tex]t_{x}[/tex]

    [tex]\frac{2000}{v_{ix}} = \frac{-v_{iy} \pm \sqrt{v_{iy}^2-4*4.9*yi}}{-g}[/tex]

    Substitute my values for [tex]v_{iy}[/tex] and [tex]v_{ix}[/tex] and [tex]y_{i}[/tex]

    [tex]\frac{2000}{cos45*v_{i} } = \frac{-(sin45*v_{i}) \pm \sqrt{(sin45*v_{i})^2-4*4.9*500}}{-9.8}[/tex]

    Now if I could only knew how to extract [tex]v_{i}[/tex] out of this mess, I'd have my answer.

    How do do that, or...

    Is there an easier way to tackle this problem?
     
    Last edited: Feb 21, 2005
  2. jcsd
  3. Feb 21, 2005 #2
    Hmm this would be.
    [tex]t_{x} = \frac{v_{ix}}{d_x}[/tex]
    so that would mean?
    [tex]t_{x} d_x = v_{ix}[/tex]
    Remember [itex] v*t = D[/itex]

    [tex]
    t_{y} \neq{v_{iy}+v_{iy}t-\frac{1}{2}gt^2 \neq 0
    [/tex]
    [tex]
    {x_{iy}+v_{iy}t+\frac{1}{2}gt_y^2 =x_{yf}
    [/tex]
    What does [itex] x_{yf}[/itex] equal?
     
    Last edited: Feb 21, 2005
  4. Feb 21, 2005 #3

    tony873004

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    Thank you for catching that. I knew with all the TEX I'd slip. It should read
    [tex]x_{f} = {v_{iy}+v_{iy}t-\frac{1}{2}gt^2=0 [/tex]

    But I don't think the error followed me to my next line, where I made a quadratic equation out of it.

    And you're right. I inverted my v and d. But that will only invert the left side of my equation and leave me with the same problem.

    Thanks for catching those. I'm going to edit the post. Any ideas on how to solve it.
     
  5. Feb 21, 2005 #4
    The [itex] t = \frac{D}{v}[/itex] followed again a few lines down.
    Also your corret equation is still wrong:
    The units of this would be
    [displacement] = [velocity]+[displacement]+[displacement]

    [tex]x_{f} \neq {v_{iy}+v_{iy}t-\frac{1}{2}gt^2 \neq 0 [/tex]

    The final [itex]x_{fy}[/itex] position does not equal zero that would mean that it did not move at all.

    This is the equation I think you are getting at:
    [tex]{x_{iy}+v_{iy}t+\frac{1}{2}gt_y^2 =x_{yf}[/tex]
     
  6. Feb 21, 2005 #5

    tony873004

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    [itex] t = \frac{D}{v}[/itex]
    I changed that one too. I believe it is in the correct order after my last edit.

    the line
    [tex]x_{f} = {v_{iy}+v_{iy}t-\frac{1}{2}gt^2=0 [/tex]

    should have read
    [tex]y_{f} = {y_{i}+v_{iy}t-\frac{1}{2}gt^2=0 [/tex]

    Thanks for catching that. I believe it was just a typo though, as I don't believe it follows me into my next line of the quadratic equation.

    I got this equation by rewriting one in the book
    [tex]x=x_{0}+v_{0}t-\frac{1}{2}gt^2=0[/tex]

    Thank you for your help so far :smile:
     
    Last edited: Feb 21, 2005
  7. Feb 21, 2005 #6
    No problem on the helping part. I am pretty sure though that:
    [tex]y_{f} = {y_{i}+v_{iy}t-\frac{1}{2}gt^2=0 [/tex]
    will not equal zero. This is becasue [itex]y_{f}[/itex] can not equal zero when the chemical retardant reaches the fire. The cannon is 500 metters above the house.
    "The cannon is 500 meters above the height of the fire."
    That means [itex]y_f = -500 \ {When \ g \ is \ considered \ negative \ or \ } y_f = -500 \ {when \ g \ is \ considered \ posative} [/itex]
     
  8. Feb 21, 2005 #7

    tony873004

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    But y initial, the height of the cannon was considered 500m. I didn't use 500 until I started plugging into my quadratic. But if the cannon is 500 then the ground should be 0. Right??
     
    Last edited: Feb 21, 2005
  9. Feb 21, 2005 #8
    That is how I would do it. It ussaly simplifes things to consider the starting position as the zero point. You do not have to do this if you are unconfortable with it.

    more to come in a moment

    edit:
    ahh yes I did not notice your edit untill now.
     
    Last edited: Feb 21, 2005
  10. Feb 21, 2005 #9

    tony873004

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    I'm copying an example in the book where a balloon is rising at a constant speed, and someone drops something. The balloon is 20m high at the time, and the example uses 20 as y initial, and 0 as the ground. More to come?... :smile:
     
  11. Feb 21, 2005 #10
    These types of problems are usually easier to solve if you break of the y component into two parts.

    1. Find the time and the displacement for where the projectile reaches the top of it's arc. The top of the arc is where a = 0 so finding the time it takes is easy. Once you know the time it is simple to find the height change.

    2. Find the time it takes for the projectile to drop from its maximum height to its final height. This is simpler since the equation becomes:
    [tex] x_f = x_i-\frac{1}{2}g t^2[/tex]

    Breaking the problem up this way adds steps but each step is much simpler then the one you are trying to solve.

    does this help?
     
  12. Feb 21, 2005 #11

    tony873004

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    I agree with you that it's easier to break it up into 2 steps if time was what I was after, and I was given either an initial velocity and an angle, or the y-component of the initial velocity.

    But the tricky part of this problem comes from the fact that initial velocity is not given in any form, and velocity is what we must compute.

    That's why I constructed it as a simultaneous equation solution. But I don't know how to split half of a simultaneous equation into 2 parts, so I used the longer method that involves creating a quadratic equation so it can be solved all in one step. Also, I'm pretty sure that in the +/- part of the quadratic, that I'm going to discard the + and keep the -.

    I appreciate your help. We get 12 problems on our homework, but he only grades one problem. Do 1 problem, but do the one he grades give you a +, but do 11 problems and skip the one he grades gives you a 0. And he told us today that the graded problem would be a projectile problem. This is one of 3 projectile problems, and the only problem out of the 12 that is giving me problems.
     
    Last edited: Feb 21, 2005
  13. Feb 21, 2005 #12
    blah... I thought of a better way find the time on the x axis then plug that into the equation for y distance.

    [tex]
    t_x = \frac{2000}{v_{ix}}
    [/tex]

    [tex]
    x_f = x_0 +v_{iy} (\frac{2000}{v_{ix}}) -\frac{1}{2}g(\frac{2000}{v_{ix}})^2
    [/tex]

    viy=vix=vi

    solve for vi
     
    Last edited: Feb 21, 2005
  14. Feb 21, 2005 #13

    tony873004

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    viy^2 +vix^2 = vi^2

    so I could re-write that as: (change the Viy to Vix since they equal each other)
    [tex]x_f = x_0 +v_{ix} (\frac{2000}{v_{ix}}) -\frac{1}{2}g(\frac{2000}{v_{ix}})^2[/tex]

    give me a few minutes to look this over.
     
    Last edited: Feb 21, 2005
  15. Feb 21, 2005 #14

    tony873004

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    since [tex]x_{0}=0[/tex] and in [tex]v_{ix}(\frac{2000}{v_{ix}}) [/tex]the [tex]v_{ix}[/tex] 's should cancel, and [tex]x_{f}=2000[/tex], I'm left with:
    [tex]2000 = 2000 - \frac{1}{2}g(\frac{2000}{v_{ix}})^2[/tex]

    but only 2000 - 0 = 2000, and infinity is the only number I can plug in for Vix to make everything to the right of the minus sign equal 0.

    Did I do this right? Algebra is not my strong point :yuck:
     
  16. Feb 21, 2005 #15
    i was puting xf and xi but I ment them only in genral terms they are yf yi. Sorry for not using consisant notation. yi = 500 yf =0 with the way you have the problem set up. So:
    [tex]0= 500 +2000 - \frac{1}{2}g(\frac{2000}{v_{ix}})^2[/tex]
     
  17. Feb 21, 2005 #16

    tony873004

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    If I did my algebra right, vix comes out to 88.5, which seems reasonable. Stay tuned... I'm going to try this a different way and see if they're the same. back in 15m!
     
  18. Feb 21, 2005 #17
    Easiest way to cheak is to plug the numbers back into the equation. Find the time taken for the projectile to get to the house and then plug the time and the velociy in the y euation to to make sure to gives the projectile reaching zero meters.
     
  19. Feb 21, 2005 #18

    tony873004

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    That makes sense. Let me try that. I'll show you my more complicated method of checking in a few minutes.
     
  20. Feb 22, 2005 #19

    tony873004

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    Your method of checking seems to work. I get 22.58 seconds now that backwards checking is simple kinmatics. 22.58 * 88.5 = 1998.33, close enough to 2000 to blame on rounding errors.

    Thanks. Sometimes I check these kind of things by writing simple computer programs to numerically integrate the problem and break it into a few thousand steps, updating x,y, and vy using g with each step. In this case, I would have started plugging in guesses for Vi, and zeroing in on it until my x value equaled 2000. But I don't have to do that now. I didn't quite follow the step where you came up with the magic formula that simplified the whole thing
    [tex]0= 500 +2000 - \frac{1}{2}g(\frac{2000}{v_{ix}})^2[/tex]
     
  21. Feb 22, 2005 #20
    Ok no magic here
    [tex]
    t_x = \frac{d}{v_x}
    [/tex]
    and
    [tex]
    x_{fy} = x_{iy} +v_{iy} t_y -\frac{1}{2}g*t_y^2
    [/tex]
    But we know that [itex] t_y = t_x [/itex] the projectile can only take one time to get to the house so they must be equal to each other. so
    [tex]
    x_{fy} = x_{iy} +v_{iy} t_y -\frac{1}{2}g*t_y^2 = x_{iy} +v_{iy} t_x -\frac{1}{2}g*t_x^2
    [/tex]
    but we know:
    [tex]
    t_x = \frac{d}{v_x}
    [/tex]
    Therefore:
    [tex]
    x_{fy} = x_{iy} +v_{iy} t_y -\frac{1}{2}g*t_y^2 = x_{iy} +v_{iy} (\frac{d}{v_x}) -\frac{1}{2}g*(\frac{d}{v_x})^2
    [/tex]
    Where d is the horizontal displacement
    Does that make sence?
     
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