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Projectile Motion (shooting an arrow)

  1. Oct 29, 2007 #1
    1. The problem statement, all variables and given/known data

    An arrow is shot at 42 degrees from the horizontal to strike a target that is 53 meters away and at the same elevation as the archer. What is the initial velocity?

    2. Relevant equations
    I'm not sure which equation(s) is/are needed so I'll list all the ones I can...
    (-Vi)(SinX) = (Vi)(SinX) + (g)(t)
    (Vfy)^2=(Viy)^2 + 2(g)(y)

    g=-9.81 m/s

    3. The attempt at a solution
    I don't know how to solve it. All the information you are given is theta = 42 and the X distance is 53 m, I don't see how you can use any of those equations to solve for anything with only those two pieces of information.
  2. jcsd
  3. Oct 29, 2007 #2


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    The only equations you should need are the kinematic equations which are given here:


    The trick with these problems is to treat the two components of motion separately. You know the x component of velocity will be [itex] u_x=u\cos \theta[/itex] (I've used u for initial velocity because v is usually used for final velocity) and I think you've identified that the distance will be given by [itex] s=u_xt [/itex] (where I've used s in place of your x). All you need to do is find an expression for t which is where the vertical motion comes into it.

    What can you write for the vertical motion?
  4. Oct 30, 2007 #3

    X= Vix (t)

    so when you plug in numbers and solve for t you get..

    t= (Vi*Cos42)/(53)

    then if you substitute that equation for t into


    you get


    which equals 9.7

    I'm a little worried thought because the answer my teacher gave me was 8.85

    is that close enough, maybe a little off because of rounding, or is it just wrong?
  5. Oct 30, 2007 #4
    The equation below is right, so if you want to solve for t, it will be, t=X/Vix not Vix/X
    X= Vix (t)

    So, there is a mistake in this equation below..
  6. Oct 30, 2007 #5


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    As pinkyjoshi points out you have calculated t incorrectly. Even more perplexing is the fact that I get a different answer to your teacher.
  7. Oct 30, 2007 #6
    Thats not good...not good at all.
  8. Oct 30, 2007 #7


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    Ok so now you have [itex] t = \frac{s}{ucos \theta}[/itex] and you've subbed it into:

    [tex] \frac{2usin \theta}{g} = t [/tex]

    If you rearrange in terms of u and plug the numbers in you do get a different value from the one your teacher says is correct. I suspect they've made an error or given the answer to something else etc.
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