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Projectile Motion help.

  1. Jan 11, 2005 #1
    I've got a question on projectile motion and don't know where to start.

    A dart player stands 3m from the wall on which the board hangs and throws a dart which leavs his hand with a horizontal velocity at a point 1.8m above the ground. The dart strikes the board at a point 1.5 m from the ground. Calculate

    a) The time of flight of the dart
    b) The initial speed of the dart
    c) The velocity of the dart when it hits the board.

    The problem is at a) and i think once i know a) i should have no problems doing the rest. Thanks alot.
  2. jcsd
  3. Jan 11, 2005 #2
    vertically the dart drop 0.3 meter.... you know
    [tex] x(t) = x_{y0}+v_{yo} t + 1/2 g t^2 [/tex]
    what is the time.....
  4. Jan 11, 2005 #3


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    What are your ideas to solving the problem...??What do you need to know??

  5. Jan 11, 2005 #4


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    There are 2 components of the motion to consider. The horizontal motion which is at constant speed (neglecting air resistance) and the vertical motion which undergoes a constant downward acceleration due to gravity.

    Consider a Cartesian coordinate system (an [itex](x,y)[/itex] plane) where the dart starts at the origin and its position at time [itex]t[/itex] is [itex](x_t,y_t)[/itex]. Let [itex]v[/itex] be the initial horizontal velocity when the dart is released.

    Then the equations of motion for the dart are :

    [tex]x_t = vt[/tex]

    [tex]y_t = -\frac{1}{2}gt^2[/tex]

    (Can you see why ?)

    Now figure out what [itex]y_t[/tex] should be at the point when the dart hits the board, and solve for [tex]t[/tex] using the second equation to get the first answer.
  6. Jan 11, 2005 #5
    My ideas so far are, to find an equation which has no variable, time,

    [tex] V_y=V_o_y - gt [/tex]

    But this would not be useful as I don't noe the initial and final velocities in the y-axis. But using displacement, which I have alot of information on, I would get,

    [tex] y= V_o_y t - \frac {1}{2} gt^2 [/tex]

    just sub ing in y= 0.3, i would have 2 unknowns which is [tex] t [/tex] and [tex] V_o_y [/tex] and I can't seem to find another equation which contains these 2 unknowns as well.
  7. Jan 11, 2005 #6


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    It would have been nice and useful to him,had he figured it out all by himself and not have his problem solved by someone else..;

  8. Jan 11, 2005 #7
    the dart is shooting horizontally, what do you think the [tex] V_{oy} [/tex] is

    our spirit here is not to give the answer

    did you ready the sticky, if don't, read it here

    the quote above is a mistake.... here you go
    Last edited: Jan 11, 2005
  9. Jan 11, 2005 #8
    Yes, curious i do understand why it is [tex] x= V_x t [/tex] but not really [tex] y= -\frac{1}{2} gt [/tex]. Why do we assume [tex] V_o_y [/tex] to be 0? Doesn't it leave that guy's hand at a certain starting velocity?

    edits: to vincent: Hmmm, ok, i did not read the part where it said "horizontal carefully". Lol i understand the rules in homework help. In fact, I've been trying to solve this for the past hour, and can't get started at all.

    Thanks for the help, I should be able to handle the rest......

    : )
    Last edited: Jan 11, 2005
  10. Jan 11, 2005 #9


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    Ummm...I hardly think I gave away the answer. I explained the reasoning, gave basic equations that can be found in any introductory text, and let him do the rest. If I had plugged in the values and given the answer, then I can understand your bellyaching, but this was hardly the case.
  11. Jan 11, 2005 #10
    gaving out the equation has no different with plug in the number for him, you should let him think what formulas should he use... read my qoute above......
    or here
  12. Jan 11, 2005 #11


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    You gave essentially the same formula I did (except yours is illucid) AND you gave away what value to use for the drop in height (even though computing that would be trivial). In essence, you "gave away" more than I did.

    I reckon I taught the orig. poster more about how to approach the problem because I explained the principles at work (no air resistance, gravity). You did not.

    Look, arguing like this is pointless. It's a matter of personal judgement what "helping too much" is anyway. The only people I am against helping are the ones that expect to be spoonfed. This poster was obviously not one of those, because he even said he could do the rest of the parts himself.
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