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Super Fun Projectile Motion Problem! Such Wow

  1. Nov 25, 2013 #1
    The Problem:

    A plane is dropping supplies to a clearing in the forest.
    The plane will approach the clearing at an altitude of 150 m while flying level at 40 m/s. Why can't the plane drop the supplies while directly over the clearing and at what distance should it release them to arrive to its destination?

    I've worked out 2 answers and I'm not entirely sure which is the correct one.

    Relevant Equations:

    d=vt+1/2at^2

    d=1/2at^2

    d=vt


    The Attempt:

    Solution 1

    d=vt+1/2at^2

    0=(40 m/s)t + 1/2(-9.8m/s^2)t^2

    0=t (40 m/s - 4.9 m/s^2 t)
    40 m/s = 4.9 t
    t= 8.2 s

    d=(40 m/s)(8.2 s)
    d=328 m


    Solution 2

    d=1/2at^2
    150 m=4.9 t^2
    t= 5.5 s

    d=vt
    d=(40 m/s)(5.5 s)
    d=220 m

    Something is telling me I'm wrong/approaching this incorrectly/way off. So please, some one...help!
     
  2. jcsd
  3. Nov 25, 2013 #2

    SteamKing

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    Solution 1 would be correct only if the supplies were shot vertically upward out of the plane with an initial velocity of 40 m/s. You can't just plug numbers into a given formula and expect the formula to check that their physical interpretation is correct. YOU have to do that before selecting and using the formula.
     
  4. Nov 25, 2013 #3
    I like your input but could you elaborate a little more? I'm doing an online course so I am on my own a lot, any advice would be great.

    I don't know what other equation to use in this case. Should I attempt to find the angle? I see no other way to solve this.
     
  5. Nov 25, 2013 #4

    SteamKing

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    Velocities and accelerations have a direction associated with their magnitudes. Obviously, if you throw an object upward with a certain velocity, it is going to take longer to hit the ground than if you simply dropped it.

    For your airplane problem, the plane is flying over the intended target with a horizontal velocity of 40 m/s. Anything dropped out of the plane will have this same horizontal velocity, neglecting the effects of drag. At the same time that the object is dropped from the plane, it begins to accelerate in the vertical direction as the force of gravity pulls it earthward. If the object takes 2 seconds to hit the ground, for example, it will have traveled a horizontal distance of 40 m/s * 2 s = 80 m from the drop point. For these types of projectile problems, the horizontal motion is calculated separate from the vertical motion, because there is accelerated motion in the vertical direction due to gravity, while the horizontal motion occurs at constant velocity.
     
  6. Nov 25, 2013 #5
    So then should I be trying to find the initial velocity then? I'm not given any angles do I don't see how to go about doing that. My intuition tells me that I should be determining time but my answers seem wrong. Any advice?
     
  7. Nov 25, 2013 #6

    SteamKing

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    Unless you are given initial velocity and direction, assume that the velocity is zero.

    For your airplane drop problem, you are given the airplane's altitude and velocity. You know based on the altitude of the plane that a certain time will elapse before the supplies reach the ground. During this time, the supplies are traveling horizontally at the same velocity as the plane. The distance from the clearing at which the drop should be made is equal to the time it takes for the supplies to fall vertically a distance equal to the plane's altitude, multiplied by the speed of the plane.
     
  8. Nov 25, 2013 #7
    Alright, I'm with you so far, I hope;

    Here's my new attempt:

    d_v=1/2at^2
    -150=-4.9 t^2
    t=6 s

    d_h=vt
    d_h=(40 m/s)(6 s)
    d_h= 240 m
     
  9. Nov 25, 2013 #8
    Looks reasonably accurate - the working is nice, but why would you round the time to 1 significant figure? It makes a substantial difference to the final answer to round the time to 6sec.
     
  10. Nov 25, 2013 #9
    The way my book has been showing me it seems to round a lot of the time, specifically in cases such as this one. I too agree that it's a substantial difference, but I also don't want my answers to deviate too far from their answers/ how the textbook has taught me. I assume that it's just because it's easier to work with round numbers, but it actually bothers me when the book does this kind of rounding. Do you think I should maintain the significant decimal places, or should I do what the book does? I just don't want to lose marks, would a teacher be that petty?

    Also, on a more important note, my answer seems sound despite my rounding error? If i keep 2 sig figs the time will be 5.5s and so then (40m/s)(5.5s)=220m, a big difference just like you were saying, I may have answered my own question just now.

    Thank you so much for your guidance, I feel I'm starting to get a lot more comfortable with these concepts, I can't thank you enough!
     
  11. Nov 25, 2013 #10
    Wait, you're not SteamKing, I mistook you for him, but I will say a thanks to you as well mic* !
     
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