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Another Horizontal Displacement Question

  1. Aug 16, 2008 #1
    First off, I'm glad I stumbled upon this place. Secondly, taking physics as an online course is probably not the best judgment call I have made in recent times. However, I'm here now and I plan on sticking through with the course. My problems, however, are lying with the labs. Since they give us websites and a book, I'm attempting at piecing things together as best I can. The website they gave us for this particular lab assignment is here: http://www.hazelwood.k12.mo.us/~grichert/explore/dswmedia/range.htm

    1. The problem statement, all variables and given/known data

    This is regarding a 'no air' set up. The initial setting for the course is V= 50m/s and a launch angle of 45degrees. The time it takes for the shot to land is 7.1s. (though there is a part where the air will need to be turned 'on'. According to the site, it's based on vc, velocity and a constant *which I would assume to be gravity*)

    2. Relevant equations

    The lab wants the horizontal displacement, which, after having to poke the internet for some clues, I believe equates to D= Vi t + 1/2at^2. I've also seen something much simpler looking at d=vt.

    3. The attempt at a solution

    Utilizing the first equation,
    D=(50m/s)(7.1s)+ 1/2(0)(7.1s) *7.1 is the total time so no squaring...I think*

    Utilizing the second,

    Now, here's the catch. According to the graph supplied with the website, I'm showing an answer that is 100m too far as it lands well before the 300m mark. That and I don't see where the 45 degree angle came into play on the equation itself.

    This is just the starter question. There are various angles in which a displacement is needed but I don't understand how to get the starter one up first. Any help, please?
  2. jcsd
  3. Aug 16, 2008 #2
    I think the problem is that you haven't learned to use vectors with these equations yet. If the object is launched at 50 m/s directly upward, it won't travel any horizontal distance at all. If the object is launched 50 m/s horizontal, and travels for 7.1 seconds, it will go 355 meters. But if the angle is somewhere between 0 degrees and 90 degrees, then it will land somewhere in between. This is where vectors come in to play, since the object has velocity in both the horizontal (X) direction and the vertical (Y) direction.
  4. Aug 16, 2008 #3
    Learned to use them? They haven't even been mentioned in the book yet. I feel like a total fish out of water. The text has only so far barely even mentioned parabola.

    If I'm even remotely understanding this, the horizontal velocity will not change. So, if I assume that both the horizontal and vertical velocities are 50m/s, then the horizontal will continue to be 50m/s while the vertical will have to adjust with gravity.


    d=1/2g(or a)t^2
    d=34.79m for the vertical distance.

    Is it safe to assume that this 34.79m is the vertical vector of y?

    ...hold on. Now the numbers aren't working for me. Now I'm getting 247m and I can't seem to figure out how i got 34 just a few minutes ago. This is all entirely too frustrating.
    Last edited: Aug 16, 2008
  5. Aug 16, 2008 #4
    This lab, and all problems similar to it, are very difficult conceptually unless you've learned vectors. I don't know what sort of online course would try to have you do this lab before learning vectors, but it's not my place to judge. At any rate, there's not much we can do to help you here at PF; any further posts would probably confuse you more.

    I would suggest learning vectors before you try solving this problem, because even if you get the correct answer by accident after trying formula after formula, you won't understand why it's correct. Try this website: http://www.ddart.net/science/physics/physics_tutorial/Class/vectors/vectoc.html and there are many others like it out there.
  6. Aug 16, 2008 #5
    Pardon me while I try and sound this out.

    In the simulation, the initial launch angle was 45 degrees. The initial launch velocity was 50m/s. This took 7.1 seconds to complete.

    x: final horizontal position
    xo: initial horizontal position
    vo: initial velocity
    t: time
    (\\): launch angle (since I can't make the funny symbol it uses)

    y: final vertical position
    yo: initial vertical position

    x-xo=vox t + 1/2 ax t^2
    y-yo=voy t+ 1/2 ay t^2

    Since there's no air resistance, and gravity only affects the vertical, the acceleration for x is zero.

    x-xo = vo cos(//)t and
    y-yo = vo sin(//)t + 1/2ay t^2

    Solving for simply x would be cake (in my slightly befuddled head) since xo = 0.


    However, this gives me some insanely large number such as 1324 or some insanely small number such as 76 if I break it down and do the components individually. EDIT: I squared the time at the end when that's only done on the second half of the vert equation. Removing that gives me 186.4

    In order to combine the two equations, we need a common something. It looks like in this case, the only constant would be 't'. So...since the second half of the vertical equation is required since gravity is involved, solve the horizontal equation for 't'.

    t = (x-xo)/(vo cos(//)

    Sub that into the equation for the vert.

    y-yo = vo sin(//)[(x-xo)/(vo cos(//)] + 1/2 ay [(x-xo)/(vo cos(//)]^2

    Simplify the equation just a bit...

    y-yo = (x-xo) sin(//)/cos(//) + 1/2 ay [(x-xo)/(vo cos(//)]^2

    Plug in the known variables. Initial horizontal position was zero, so x = 0. The initial vertical position was zero, as was the ending so y - yo = 0. Acceleration was due to gravity so g = -9.8 m/s^2 (directed downward).

    This would mean that 0=x sin(//)/cos(//) + 1/2 (-9.8) [x/vo cos(//)]

    Now here's where I get confused. I need to solve for x but the example I'm looking at doesn't make much sense at this point. It moves to show that there are 2 possible solutions.

    x = 0 or x =(v^2 sin(//) cos(//))/4.9

    I can work with this equation, I suppose but I don't get how it got that way.

    x= (50)^2 sin(45) cos(45) / 4.9
    x= 228.06m

    Since x was for a distance, and the original problem was in meters, then the solution should also be in meters.

    My brain hurts and I still think I'm miles off here. If I'm thinking right, the answer should be closer to 255m if I utilize the graph.
    Last edited: Aug 16, 2008
  7. Aug 16, 2008 #6


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    This looks like the right answer to me.

    If you are getting 186.4 from the above equation, I think your calculator is set to radians (so you are calculating cosine of 45 radians instead of cosine of 45 degrees). If you set it to degrees, I think you'll get the answer you're looking for (since the problem indicates you're just looking for the horizontal displacement). What do you get?

    (The vertical equation is not needed here, because they gave you the time of flight. If they had not given you the time for the motion, they you could use the vertical equation to find the time for it to land (when y-displacement is zero). But here that information is given to you already.)
    Last edited: Aug 16, 2008
  8. Aug 16, 2008 #7
    ...you mean all this time my -calculator- was set to the wrong thing? *headdeskthud*

    Okay, I reset it to degrees and got 251.02. Which is insanely close to the approximated 255 I was looking at on the graph.

    All in all, I'm way over-thinking this problem, aren't I?
  9. Aug 16, 2008 #8


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    You took some wrong turns for this problem, but when you have a calculator error of course you'll be thinking of all kinds of reasons why things aren't working out.

    But more importantly, I think it would be much more common in these problems for them to not give you the time of flight, in which case you would have to use both equations (horizontal and vertical) together to solve the problem, in a fashion similar to your last post. So working these things out and getting them straight in your mind can only be a good thing.
  10. Aug 17, 2008 #9
    Another one of those "thinking it out" things.

    Now the "Air" has been turned on.

    I'm thinking that the equation would change slightly (since it still wants the horizontal disposition) to:

    x-xo = vo cos(//)t + 1/2 at^2

    Plug in the same set of variables (with the time changing due to the new resistance).

    x = (50)cos(45)(6.5) + 1/2(-9.8)(6.5)^2

    ...and get 22.78. Looking for something more on the lines of 180.

    x = (50)cos(45)(6.5) + 1/2((-9.8)(6.5))^2

    Getting 2258. Definitely not it.

    Since I still have the time, I still don't need the y-equation. Adding in the 1/2at^2 is the only thing I can think of to even get the resistance in there. Gravity is the only force that acts upon an projectile.
  11. Aug 17, 2008 #10
    Air resistance is difficult to analyze numerically without Calculus. If this is called a "thinking it out" question does that mean your teacher wants actual numbers? Maybe it is better to think conceptually. Use your common sense, imagine there is air resistance, and answer the following: do you expect the final landing spot of the ball to be greater or lesser than before (without air)? Will the time of flight be greater or smaller? Will the acceleration in the X-direction still be zero? Will the acceleration in the Y-direction simply be 9.8?
  12. Aug 17, 2008 #11
    No, more of a thinking it out for me kind of thing. The lab still wants the horizontal displacement.

    According to the graph provided on the website, when the air is turned "on", it uses a drag force of cv (v is the velocity and c is a constant). However, it does not give the constant. Since gravity is the only force acting upon a moving object, I would assume that the constant would be gravity.

    So, if it's cv then

    (50)(-9.8)=-490 but I don't know how to fit that into the equation.
  13. Aug 17, 2008 #12


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    I think you may have to solve the equations of motion form scratch again with the new air resistance term in it. Unless somewhere in your notes or textbook they've done it for you.
  14. Aug 17, 2008 #13
    Yeah, if this is an introductory physics course I don't see how you can be expected to solve this numerically. You can't use the kinematics equations because they assume constant average acceleration. With air resistance, the vertical acceleration is not 9.8 and the horizontal acceleration will not be zero!

    I suppose you could estimate it step-by-step, in like 0.5-second intervals, and find the new velocities after each interval, but that would be a real hassle and it sounds like you've had enough hassle. Try to contact your instructor and see how s/he wants you to approach this.
  15. Aug 18, 2008 #14


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    And the fact that acceleration won't be constant.

    Excellent advice :approve:
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