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Projectile Motion Check + Good tutorial?

  1. Apr 10, 2010 #1
    1. The problem statement, all variables and given/known data
    I have this problem, but not solutions, and my answer looks wrong but I can't really see that I've done anything wrong. So I was wondering if someone could check:

    A transport plane travelling directly at a steady speed of 50m/s at an altitude of 300m releases a parcel when directly above a point X on level ground. Calculate:
    a) The time of flight of the parcel
    b) the speed of impact of the parcel
    c) the distance from X to the point of impact

    (all assuming g = 9.8m/s^2

    2. The attempt at a solution

    So I drew a diagram. For a) i got:

    x = 300m, a = -9.8m/s^2, u = o, t = ?
    [tex]300 = \frac{1}{2}(-9.8)t^{2}[/tex]
    => t = 7.82s

    b) I got x = 300m, a = -9.8m/s^2, u = 0, v = ?
    [tex]v^{2} = 0 + 2(-9.8)(300) \Rightarrow v = 54.2 m/s[/tex]

    c) [tex]x_{h} = 7.82 * 50 = 391 m[/tex]

    Now what I'm not sure about is did I do that correctly? and can I assume that initial velocity (u) is 0? And then the equations, they require me taking a square root of a negative number (which I can't do over the real number field) so can I modulate the number, or am I doing something wrong?

    Also, could anyone recommend a good projectile motion / applications of newton's laws problem solving tutorial on the net? I can't seem to find one :(

    Thanks!
     
  2. jcsd
  3. Apr 10, 2010 #2
    You did one thing incorrectly. The initial speed of the package as it leaves the plane is not 0. Think about it. What would the speed of the package have to be in order to travel along with the plane?

    Also your equations are making you take square roots of negative numbers because you're not using your coordinate system correctly. You let down be the negative direction but you said your box (which falls downwards) has a positive displacement. So you are contradicting yourself.
     
  4. Apr 10, 2010 #3
    Hmm. But with the package, since the plane is flying steadily at 50m/s doesn't that mean that the initial vertical speed is 0?

    And I see what you mean by the incorrect direction
     
  5. Apr 10, 2010 #4
    Ah yes the initial vertical speed is zero, I misread your equation since you used x's instead of y's. Sorry about that.

    So your answers are indeed correct.
     
  6. Apr 10, 2010 #5
    Ah kk, thanks!
     
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