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Combination of freefalling and constant horizontal movement

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data
    At the height of H, from a plane with the speed of V, there is an object "thrown off".
    H = 10km = 10 000m
    V = 300m/s

    To be determined:
    - total distance from the plane to the falling point - S
    - the time the fall takes - T
    do not have to take aerial friction into account
    2. Relevant equations
    s = v0t + at²/2



    3. The attempt at a solution
    -Take the positive direction of movement as movement from the ground to the sky
    -horizontal acceleration = 0 , vertical acceleration = -g = -9.8m/s²


    the total distance is the sum of the distance travelled vertically and horizontally (sum of 2 vectors that is or that would make sense atleast)

    so S² = (VT)² + (-gT²/2)²

    But I have 2 unknowns and 1 equation. What will I have to do next?

    Also I apologise in advance - I don't know all the mathematical terminology in English that well, so some things may seem a bit confusing :/

    Cheers
     
  2. jcsd
  3. Oct 15, 2012 #2
    You know the distance of vertical travel: this is the height from which the object is released.
     
  4. Oct 15, 2012 #3
    I'm sorry, that was rather vague - what exactly do you mean by that?

    Do you mean that horizontal travel lasts exactly the same time as freefall would?
    so if the time of freefall were Tf = (2H/g)^0.5 ~ 45 seconds
    Horizontal travel - 13500m

    So total distance S² = 13500² + 10000² ?, plug into the original equation and solve for T?
     
  5. Oct 15, 2012 #4
    The horizontal travel happens simultaneously with the vertical travel, as they are parts of one and the same single motion. So yes, their durations must be equal.
     
  6. Oct 15, 2012 #5

    Simon Bridge

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    This is a problem in "ballistics" (you may want to look it up) ... you treat the horizontal and vertical components of the motion separately. Draw a v-t diagram for each motion: displacement in each direction is the area under the graph and the acceleration is the slope - as voko says, the time is fixed by the vertical distance and the acceleration of gravity.

    The question seems to be asking for the horizontal distance to the landing point.
     
  7. Oct 15, 2012 #6
    I got it solved and the solutions are 96, 685s and -35,461s

    But when I took the positive direction movement the same as the falling object's, the solutions are the same only -96, 685 and 35, 461.

    With the first pair of solutions the parabole is upside down and the 96, 685 is T where S=-(13500² + 10000²)^0.5 (~-16800m)
    but what is the other T? S is still the same, but what S is that?



    EDIT: It made me think of another thing - S is only the displacement between the starting point and the landing point, how would I find out the total distance travelled by the object, because the trajectory of the object is curved hence it should be longer than the displacement.
     
    Last edited: Oct 15, 2012
  8. Oct 15, 2012 #7
    Now I am confused. What are these times? You solved the problem in #3 as far as I can tell.

    I do not know what word was used in the original language of the problem, but in English "distance" and "displacement" mean the same thing. The length of a trajectory between two points will obviously be different (except for wholly unaccelerated motion) from the distance between the points, but to find this length you need to apply calculus.
     
  9. Oct 15, 2012 #8
    I only found the time it took for freefall at the height of H in #3.
    The time, how long it travels the entire distance, I got from:
    -gT² + VT = -S and S² = H² + (Horizontally travelled distance during the freefall time)²

    96 seconds sounds very much more realistic than 45 seconds, because it's also moving horizontally - well that is how I understood it.
     
  10. Oct 15, 2012 #9
    As Simon Bridge and I explained to you earlier, these are one and the same time. The time to reach the ground from height H is INDEPENDENT of any horizontal motion; it depends ONLY on H (and initial VERTICAL velocity, if any).
     
  11. Oct 15, 2012 #10
    Wait, so if I throw a stone off a roof or I let it drop down - the time for it to reach the ground is always the same? Assuming I'm not aiming toward the sky but exactly horizontally, parallel to the ground below.

    E:Oh, ofcourse the only thing different is the speed
     
  12. Oct 15, 2012 #11
    Neglecting air resistance, yes. This is because the force of gravity is independent of velocity.
     
  13. Oct 15, 2012 #12
    Yeah, well thanks alot Simon and voko - NOW I finally get it :)
     
  14. Oct 15, 2012 #13

    Simon Bridge

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    But not in physics they don't. In physics, they have a special meaning: displacement, for example, is a vector - the magnitude of the displacement is an example of a distance.

    More importantly, if you travel in a circle with a circumference C, then you have traveled a distance of C but your displacement is zero.

    Displacement is the area under a v-t graph - if the area is below the t-axis then the displacement is negative.

    In this problem, the vy-t graph is a triangle with base T and height vy ... vy/T=g (taking "down" as positive) and h=vyT/2; the vx-t graph is just horizontal, so the acceleration is zero, and the horizontal distance is d=vxT

    This gives you three equations and three unknowns.
    1. the vertical speed on arrival
    2. time to the target
    3. horizontal distance to the target

    If you solve for time you will get two answers! That is an artifact of the math used - one of the times is non-physical. (We've actually left some information out.) As usual the math contains more than what is available in the Universe, which is why we do experiments: just to check.

    That's the "ahah!" moment all us educators live for :D Well done!
     
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