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Really really need help on a projecton physics problem

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data

    A missile is fired across level ground at an angle of 37.0o to the horizontal and
    remains in the air for 7.60 s. Find the initial velocity and the range of the missile.


    2. Relevant equations

    not sure

    3. The attempt at a solution

    no inkling...I know that v0 cos 37 and v0 sin 37 will eventually lead to the range, but I have no idea how to get the initial velocity.

    I even know the answer, just no idea how to do it lol.

    v= 61.9 m/s, range = 376 m

    Thanks!!
     
  2. jcsd
  3. Feb 12, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    OK you got the range and you got the horizontal speed right?

    So if the horizontal component is 61 m/s what does that mean that the initial velocity was at 37° ?
     
  4. Feb 12, 2009 #3
    thanks for the welcome! sorry for not being clear earlier, the answer for the problem is: v= 61.9 m/s, range = 376 m

    I have no idea how to get to that point. All I can glean out of the problem is the angle and time, not enough to plug into any formulas.
     
  5. Feb 12, 2009 #4
    the initial velocity appears to be 76.4, and I could get the displacement pretty quick with that info. How would you normally get initial without knowing what the final velocity was?
     
  6. Feb 12, 2009 #5
    To find the initial velocity, first set up an equation to find the y component of the initial velocity. The equation for this is Y=Yo+Vot+(1/2)at^2. In this case Vo=Vosin37 and your displacement (Y-Yo) is zero since you are starting and ending on the same level ground. Therefore your equation will look like: 0=Vosin(37)*(7.6)+(1/2)*(-9.8)*(7.6)^2. Solving this for Vo will give you 61.879 or 61.9 m/s. You then can use this to find the range which is x=Vot.
     
  7. Feb 12, 2009 #6
    thank you so much!! I have one question though, I don't have Y=Yo+Vot+(1/2)at^2 on my formula list. Instead, I have displacement = v0t + 1/2 at^2, which is essentially the same thing without an extra v0 in the beginning. What is the extra v0 for? I often see the above equation in my physics book too...thanks again.

    nm I think I get it, its just combining the cos + the regular displacement equation...
     
  8. Feb 12, 2009 #7
    Your equation is actually the exact same as mine. When you say the extra Vo do you mean an extra Yo. Either way, your displacement is the same as Y-Y0 which could just equal Y. You have it right so don't worry.
     
  9. Feb 12, 2009 #8
    thanks!!
     
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