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Homework Help: Finding The Initial Velocity & Angle

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data

    An arrow shot into the air has a time of flight of 6s and a range of 210 meters. Find the: A.) Initial Velocity and B.) The angle at which it was shot.

    2. Relevant equations


    3. The attempt at a solution
    I don't know how to solve this.
  2. jcsd
  3. Jan 30, 2012 #2
    Welcome to PF!

    Try using the information you have to form two equations with said information in them i.e time of flight T and the range Sx.
  4. Jan 30, 2012 #3
    Did you draw your sketch?

    You can start by labelling some points. The time of flight is 6s. What does that mean? It sounds smarmy, but it's a good question to answer.

    What does the arrow's flight look like at 0s?
    What does the arrow's flight look like at 3s?
    What does the arrow's flight look like at 6s?

    What do we know about the horizontal velocity and the horizontal acceleration?
  5. Feb 1, 2012 #4
    Thanks! I think got it.

    Given: R= 210m; ttotal= 6s; tmax = 3s since ttotal = 2(tmax)

    ymax = Voy^2/2g
    = (29.4m/s)^2/2(9.8m/s)^2
    = 44.1m

    Voy = gtmax
    = 9.8m/s^2(3s)
    = 29.4m/s

    Vox = R/ttotal
    = 210m/6s
    = 35m/s

    The angle at which it was shot:

    theta = tan^-1 (29.4m/s / 35m/s)
    = 40.03 degrees

    The Initial Velocity:

    Vo = sqrt Vox^2 + Voy^2
    = sqrt (35)^2 + (29.4)^2
    = 45.71m/s

    Is this it? :D
  6. Feb 1, 2012 #5
    Use trig:


    x(any distance since speed is constant in x direction)=(V0,x)t=V0(costheta)t
    and y=V0(sintheta)t-1/2gt^2
  7. Feb 1, 2012 #6
    Yes, yes and...YES! :)

    However, I feel I REALLY need to give you a bit of warning. It can be tempting to try and memorise (or in some other fashion, remember) the derivative equations (eg Viy = sqrt (2gh) ). I would very strongly recommend against that. Not only are they only useful in specific circumstances, it is easy to forget part of it. I would instead recommend that you fall back upon your basic three kinematic equations:

    d= Vi*t + 1/2 a*t^2
    Vf = Vi + a*t
    Vf^2 = Vi^2 + 2a*d

    I believe it is much easier and safer to stick to these three equations and say to yourself, "in this instance Vf = 0" or "t-up=1/2 * t-total" and do whatever algebra is necessary instead of trying to remember a specific equation to solve for that particular instance.

    I'll hop down off my soapbox now, before I sprain an ankle or something! :)
  8. Feb 1, 2012 #7
    I think it would be a better statement to say that:

    dx = Vix*t + 1/2a*t^2 and that the acceleration in the x direction is 0.

    You cannot say that cos θ is linked to the vectors in the x direction. That is only the case when the angle is measured against the horizontal. There are cases where you will use sin θ to calculate the x components (see attached image).

    This is part of the reason I believe it is important to not to try and make such qualified statements. If you automatically say Vx = Vcos θ, there are times when you will be completely wrong. The best method is learn and memorise your trig functions using adjacent, opposite and hypotenuses (hypoteni?). cos = adj/hyp, sin = opp/hyp, tan = opp/adj.

    However, having said all of that, in this particular case you are correct.

    EDIT: I hope this doesn't appear as if I am talking down to you. I tend to get 'proper' when writing and I know that it can, at times, come off that way. This is something that I've seen cause SO many problems and so much frustration, that if this can help you avoid that it's well worth it! -r

    Attached Files:

    Last edited: Feb 1, 2012
  9. Feb 2, 2012 #8
    I usually use differentials as well to describe physical phenomena in physics, however, I know there are people completely unfamiliar with that notation out there. So, for simplicity's sake, I didn't do that. As far as the trig functions, well, I thought maybe this user would get it, assuming they need to have developed a foundation for trig before entering a physics class.
  10. Feb 2, 2012 #9
    Plus I know these formulas like a piece of cake!
  11. Feb 2, 2012 #10
    you are asked to get the Initial Velocity.. what you have is the x and y component of the initial velocity.. you still need to get Vo. how do you solve that???

    Hint: use Pythagorean Theorem
  12. Feb 2, 2012 #11
    I switch back and forth. I do find it difficult not to sneak in a little conceptual differentiation (even if I avoid using the word 'derivative'). I deal with enough non-calc students that I believe I think of the concepts in this mish-mash of calc/non-calc.

    Unfortunately, and I am not pointing fingers at anyone in particular, I've found that the assumptions about prior math ability are in error often enough that I have stopped making those assumptions. I did not mean to imply you were incorrect, but lately I've been dealing with a number of students that had an instructor who taught derivative (maybe I should say massaged since 'derivative' has calculus implications) equations instead of concepts (eg equations for MaxHeight and Range). They seem to latch onto these massaged equations and use them in situations where they don't hold. I probably reacted from that more than anything. Sometimes I feel like a Kinematic-Don-Quixote! I apologise if you took anything negative out of my posts, for it was not my intention.
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