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Projectile Time variable confusion

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data
    When I'm solving problems, I'm told that the time between the vertical and horizontal components are the same. So that means that I would have to use the same time when calculating for both the vertical and horizontal components right? But then there are times when i have to use the time i found from the vertical component and multiply it by 2 to find the time for the horizontal component to see how long it is in the air. That makes sense, so doesn't that mean that the time for both the x and y components won't be the same? Since you need to multiply one to find the other. This is for those "find the horizontal range" questions. This seems like an easy concept (which it probably is), but I'm just confused on whether i have to multiply it every time i do this type of question or to leave it after finding it :(
     
  2. jcsd
  3. Oct 8, 2011 #2

    gneill

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    Staff: Mentor

    You might find the time to the apex of the trajectory, then multiply it by two to find the time for the entire trajectory (assuming that you have a symmetrical trajectory -- the landing elevation being the same as the launch elevation). The landing time will then be twice the time to apex. Otherwise I can't think of a reason to do this.

    Can you provide an example?
     
  4. Oct 8, 2011 #3
    But then sometimes you don't multiply it and just use the time you found from the vertical...
    So here's an example:

    A mortar shell is fired at 80m/s at an angle of 65 degrees to the horizontal and falls on a plain 50m below the level of the hill from which it was fired. What is the horizontal range of the shell?
    Vvi= 72.5046 m/s
    VHi= 33.8095 m/s

    d=Vit+0.5at2
    -50=(72.5046m/s)t+0.5(-9.8m/s2)t2
    0=-4.9t2 +72.5046t + 50m
    *use quadratic formula*
    t= 15.4570 or t=-0.6601
    ignore the negative t, so t= 15.4570, which is how long the shell is in the air.

    then you find the range, using v=d/t. rearrange:
    d=VHi*t
    = (33.8095 m/s)(15.4570s)
    =524.0395 m
    therefore, the horizontal range is 5.2x102 m.

    here's another question:

    A gun fires a shell with a speed of 315 m/s. Neglecting the effects of air resistance, calculate the maximum range of this gun.
    angle= 45 degrees
    Vvi= 222.7m/s
    VHi= 222.7 m/s
    av= -9.8 m.s2

    for vertical component:
    a= (vf-vi)/t
    t= (0-22.7m/s)/ -9.8 m/s2
    t= 23s

    to find the horizontal range: i multiplied the time i found before by 2
    d= Vft-0.5a(t)2
    d=(46s)(223m/s)
    d= 1.0 km

    Therefore, the maximum range is 1.0km.
     
  5. Oct 8, 2011 #4

    gneill

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    Staff: Mentor

    In your second example you found the time from launch to apex (when the vertical velocity becomes zero). That's only half the trajectory.

    The shell will land with the same speed that it was launched.
     
  6. Oct 8, 2011 #5
    yeah since it was only half the trajectory, i had to multiply the time i found for the first half by 2 to find the time for the whole thing so i can use it in the equation to find the horizontal range. but i wanted to know how come for one question you had to multiply the time by 2, and for the other one you don't even though they both ask you to find the same thing
     
  7. Oct 8, 2011 #6

    gneill

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    Staff: Mentor

    In the first question you used the trajectory equation to find the time that the projectile reached the target (landing) elevation, not the apex of the trajectory.
     
  8. Oct 9, 2011 #7
    Oh, I see. So if the question mentions the landing elevation, like the first one did with the plain, then i must use the trajectory equation and the quadratic formula, but if it doesn't state that and only to the apex, then i must find the time to the apex and then multiply by two to find the whole time for the horizontal range?
     
  9. Oct 9, 2011 #8

    gneill

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    Staff: Mentor

    In any given problem there are often several ways of approaching it and solving it. In the cases you have here you could use the trajectory equation for both. When the trajectory is symmetrical though, it is often easier to use the time to apex simply because it avoids the quadratic!
     
  10. Oct 9, 2011 #9
    oh, I see. Thanks!
     
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