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Projectile motion of a Missile relative to ground

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data

    A missile is launched from a fighter jet cruising horizontally at a speed of 280 m/s, at an altitude of 1800 m. The initial velocity of the missile relative to the fighter jet is 160 m/s, in the forward direction. The missile is powered by its own engine, which provides it with a constant, horizontal acceleration of a = g/3. The vertical motion of the missile, however, is a free fall (as we neglect the effect of air friction). Take g = 9.80 m/s2.

    (a) How long does it take for the missile to reach the ground?
    (b) If the missile hits a tank on the ground, how far ahead of the plane horizontally was the tank when the missile was launched? Assuming that the tank has negligible speed compared with the missile.
    (c) What is the speed of the missile just before hitting the tank?
    (d) What angle does the velocity of the missile make with the ground just before it hits the tank?



    2. Relevant equations

    Δx = Vixt x 1/2gt^2

    Vf = Vi + at


    3. The attempt at a solution

    I completed this whole problem then checked the answer key and it seems that my professor made gravity = 9.80 m/s^2 positive and i made gravity negative. I just wanted to know why and b/c i have a exam tonight and want to know when to make gravity positive in these types of questions.
     
  2. jcsd
  3. Oct 5, 2011 #2

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    Conventionally, standard gravity g is always positive 9.8 m/s2 (well, 9.80665... m/s2 depending on how much precision you want). The point is that g is a positive constant.

    When substituted into an equation, the number might end up being negative, but note that g itself is still positive.

    For example, take the formula (one that you listed),

    vf = vi + at.

    Now suppose that you have a problem somewhat like the one you gave above with the missile.

    *You* get to define which direction is associated with positive or negative. Typically, up is defined as positive and down is defined as negative (although it doesn't have to be, this is your choice). So let's use that. Up is positive. Since the acceleration due to gravity points down, that That means that it points in the negative direction, such that a = -g. Our equation becomes,

    vf = vi - gt.

    But note that g itself is still a positive value. The acceleration a can be negative, even though g remains positive.

    One last substitution shows that this is

    vf = vi - (9.8 [ms-1])t.

    But don't forget that the 9.8 ms-1 isn't negative because g is negative, but rather it's negative because the formula contained a "-g" in it. g is and always was positive.

    All that being said, the direction of whether up is positive or down is positive is really your choice. But once you make that choice, ensure you are consistent with it throughout the entire problem, and make sure your choice applies not just to acceleration but also to velocity and displacement. As long as you are consistent, you'll get the same answer either way.

    If your professor has a preference (as to whether up is positive or negative) I suggest following it just to keep things simple. But whatever the case, standard gravity g is always a positive constant.
     
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