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Projectils and Trajectories

  1. Oct 2, 2005 #1
    I have a problem. So far, I have decided that this problem is using the Range Equation. Here it is - The initial speed of a cannon ball is 200m/s. If it is fired at a target that is at a horizontal distance of 2 km from the cannon, find A) the two projected angels that will result in the hit and B) the total time of flight for each of the two trajectories.


    Okay, now here is what I have so far------

    R=(Vi^2sin2theta)/g

    R= 2km
    Vi= 200m/s
    Assuming a= 9.8m/s^2

    take the eq from above and plug in the numbers and get
    2km=(200m/s)^2*(sin2theta)/9.8m/s^2

    solving for sin
    (2km) *(9.8m/s^2)/200m/s = sin2theta


    Am I correct? Please advise. If there is a better approach to this problem, please share.
     
  2. jcsd
  3. Oct 2, 2005 #2

    hotvette

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    Where did R=(Vi^2sin2theta)/g come from?

    Also, don't quite know what is meant by:

    What 2 angles? Unless I'm mistaken, this is a classic projectile problem with an initial velocity in 2 directions.
     
  4. Oct 2, 2005 #3
    The equation R=(Vi^2sin2theta)/g is the Range formula in my Physics textbook. The way I understand the Two projected angle portion of the problem, is that you will have an angle above 45 degrees and one below 45 degrees (if I understand correctly that 45 degrees is always max Range). I hope that this will shed a little more light on the problem. I am still lost.
     
  5. Oct 2, 2005 #4

    hotvette

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    Wow, never saw it like that before. Anyway, here's what I would do:

    1. You have an equation that expresses the range (i.e. horzontal distance) as a function of [itex]\Theta[/itex] and [itex]t[/itex].

    2. You should also have an equation that expresses the height (i.e. vertical distrance) as a of a function of [itex]\Theta[/itex] and [itex]t[/itex].

    3. You know the range at impact. Thus you have 2 equations with 3 unknowns ( [itex]\Theta, y, t[/itex]).

    4. But, what do you know about y at impact? This should allow you to solve the equations for [itex]\Theta[/itex] and [itex]t[/itex]. Hint: solving a quadratic equation is probably needed.
     
  6. Oct 2, 2005 #5
    There is no y Impact.... or am I reading to far into the problem? I am given v=200m/s, x=2km, (assume) g=9.8m/s^2. Now to set up for ther quadradic
    -b*sq ^2-4ac/2a
    where b=v =200m/s
    where a = x = 2km
    where c = g =9.8m/s^2

    Am I on the right track or should g=9.8 m/s^2 not be in the equation at all?
     
  7. Oct 2, 2005 #6
    The range equation is derived for a situation where you're given vi , [itex]\theta[/itex] , and xf (the range.) It also assumes that yi equals yf, as in this problem.

    The problem with working the range equation backwards (as tralblaz has in this case) is that it will give you only one of the 2 possible solutions. This is because if [itex]\theta[/itex] = 75 (for example), sin(2[itex]\theta[/itex]) = sin 150 = 0.5 = sin 30, so your calculator will tell you 2[itex]\theta[/itex] = 30 when you solve the problem. If you're aware that the 2 angles are complimentary, you're set. :wink: If not, you'll have to find another way.

    I think hotvette's suggestion to solve the problem using the 4 kinematics equations is really a better (if harder) way to solve the problem. It's much more general, so you can use that method in places where the range equation can't be used. Learning how to do it on a fairly easy problem will help you because later on you may have no choice but to solve a problem using the kinematics equations.
     
  8. Oct 2, 2005 #7

    hotvette

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    I think I see now. Forget about what I said. I'm not going to take the time to verify the math, but I suspect your range equation represents the horizontal distance at the completion of the trajectory. I was referring to the basic equations of a trajectory that can be solved to come up with the equation that you seem to already have. In that case, just solving your range equation for [itex]\Theta[/itex] provides the answer for the first part. Seems too easy.

    For the second part, I don't know what 2 trajectories are being referred to. Also, to find the time, you need an equation that relates the horizontal distance to [itex]\Theta[/itex] and [itex]t[/itex]. Since you already solved for [itex]\Theta[/itex] in part 1, you have a single equation in t that you should be able to solve w/ algebra.
     
  9. Oct 2, 2005 #8
    For a given range, R < Rmax there are 2 different firing angles that will land the projectile in the same place. If you fire a shell at 30 degrees, it will land at exatly the same spot as if you fire it at 60 degrees. The 2 angles are related to each other by [itex]\theta_{2}[/itex] = 90 - [itex]\theta_{1}[/itex], so at Rmax, [itex]\theta_{2}[/itex] = [itex]\theta_{1}[/itex] = 45 degrees

    I'm going to have to go back on my last answer. I set up a quadratic and solved for t using the kinematics equations. Unfortunately, it was degenerate (one solution being zero) and the non-zero solution produced the same result for [itex]\theta[/itex] as the range equation. Go ahead and solve it using the range equation. You just have to recognize that [itex]\theta_{2}[/itex] = 90 - [itex]\theta_{1}[/itex]
     
  10. Oct 2, 2005 #9
    Thanks!!! THat helps! By the way, I solved for t by using Xf=Xi+Vxit+½axt^2
    where xf=2000m (2km)
    where 1/2(9.8m/s^2)t^2
    solve for t
    xi and vxi are zero because thst is the initial
    so now you have t^2= 2000m/(1/2)*9.8m/s^2 which is 20.2s

    Hopefully I did the math correctly. Now if I am doing the Range portion right, here is what I have -
    R=vi^2sin2theta/g
    I know Range = 2000m
    I know vi=200m/s
    I know g=p.8m/s^2
    so now I solve for sin - (2000m)*(9.8m/s^2)/(200m/s)^2 and I get 4.9
    Now I do not know how to take 4.9 and get sin2theta. My answer when I finish plugging into the equation looks like this

    sin2theta = 2000m*9.8m/s^2/(200m/s)^2

    sin2theta = 4.9

    how do I get my angle from that (provided I did the math correctly)
     
  11. Oct 2, 2005 #10
    You're way off on the first part. You definitely shouldn't have acceleration due to gravity in your x equation. It should end up taking the final form:

    [tex]
    x_{f} = v_{i}cos(\theta)t
    [/tex]

    You're going to be better off (I've decided) solving for [itex]\theta[/itex] using the range equation.

    As for the range equation, it looks like you're off by a factor of 10. For 2000*9.8/200^2 I get 0.49, not 4.9. As for where to go from there, you just take the arcsin (sin-1) of the right side:

    [tex]
    sin (2 \theta) = 0.49 \Rightarrow 2 \theta = sin^{-1} (0.49) \Rightarrow \theta = \frac{sin^{-1} (0.49)}{2}
    [/tex]

    This will give you one of your 2 angles. Use the relation I posted above to find the other, then it's simply a matter of substituting the angles into one of the kinematics equations to solve for t.
     
  12. Oct 2, 2005 #11
    Thanks a lot!! I am glad that there are people out there that can help you out and double check your math. I am slowly picking this up. . Thanks again!!
     
  13. Oct 2, 2005 #12

    hotvette

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    Thanks Grogs - I didn't know that. Gosh, I guess I'm older than I thought. I've never even heard of a range equation. We just did everything from the very basics on every problem (actually, it's very good training) - construct FBD, write [itex]F_i=ma_i[/itex], solve the differential equations, appropriately apply the information given in the problem statement, and solve for for what you want using sound logic and understanding of the physical problem at hand. All we ever needed to know was a few basic equations, and I still remember them after all these years (but I won't tell you how many years that is!) :approve:
     
    Last edited: Oct 2, 2005
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