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Homework Help: Determining displacement using components

  1. Jul 6, 2008 #1
    1. The problem statement, all variables and given/known data
    A golfer drives a golf ball 214 m [E] off the tee, then hits it 96 m [28 degrees N of E], and finally putts the ball 12 m [ 25 degrees S of E ]. Determine the displacement from the tee needed to get a hole-in-one using (a) vector scale diagram and (b) components. Compare your answers.

    2. Relevant equations
    a^2 + b^2 = c^2
    c^2 = a^2 + b^2 -2abcosC

    3. The attempt at a solution

    Okay, well I managed to solve the vector scale diagram. Displacement ended up being 3.1 x 10^2 m. However, my problem is with the components method since my teacher didn't explain it at all, whatsoever. I've been looking through the text book trying to teach myself, and I get answers in the 300 range, but I don't know if I'm doing it correctly.

    I set up my diagram like so: 96 m is head-to-tail with the 12 m, and I drew right angles for both starting from where they meet. I find the difference in the y's, and the x's, and then I substitute them in the Pythagorean Theorum (that's what they were doing in the book?) and I get approximately 87 m, and to that, I add the initial 214 m.

    Any ideas as to where I'm going wrong? Any help is sincerely appreciated!
  2. jcsd
  3. Jul 6, 2008 #2
    Your attempt at a solution is good. The relevant equations are true, but maybe will just make you do lots more work than you need to.

    The golf shots are position vectors. The resultant is obtained by adding them, tail to head, exactly as you describe. Simply drawing the map trail of the hits is the obvious thing. From the start point, horizontal left to right, 240 long, going due East, is the first line. Its component is all E-W. The N-S part is zero.

    Now at the end point of that drive arrow, lightly draw a N-S axis line, and a E-W one as well. Add on the 96m hit, making it point 28 degrees Northward of the E-W axis. You can already see the right triangle that will tell you how much further the ball has progressed East, and from the same hypotenuse, how much it has moved North.

    So press on. At the head of the 96m vector, again lightly sketch some short helpful N-S and E-W axis lines, and draw in the path of the last 12m long hit, this time turning the vector to point 25 degrees South of the E-W axis. We leave you to put in the resultant of all those vectors added together.

    By now, you can maybe appreciate that all the E-W and N-S components can be figured from the hypotenuses, and simple trigonometry much easier than the equations you gave. The first hit involved a zero angle. cos(0)=1, so that part is easy. You only need sines and cosines. I leave you to figure which delivers the E-W components, and which the N-S components. Once you know the final position in terms of its added-up components, you then get to use Pythagoras theorem to get the length of the hole-in-1. By then you will have also seen how trigonometry equations are in the end proved by using Pythagoras.

    You need some feedback to know that you are succeeding, so I can tell you that I got 87.76m for the E-W component of the 96m hit. There was a N-S component that also happened. Then after that, the 12m hit made two components, one for E-W and another N-S. I figured the value for one of them was -5.07m. you can figure out the other. I got a value for one of the final resultants as 39.99m, the other for you to figure out. For the hole-in-one hit, you had better get an answer bigger than 340m. You can let us know by how much.

    Tip: Keep clear in your mind where to subtract. "Adding" a component can involve a subtraction depending on the direction. You need to decide when an angle is "negative".
    Last edited: Jul 6, 2008
  4. Jul 6, 2008 #3
    H'okay. I had another crack at it, and I figured out what I was doing wrong ... well, partially, because I'm not getting an answer above 340 m, and I'm also not getting the angle on the final displacement.

    For the 96 m hit, my x-component was 84.76 m, and y-component was 45.07 m.
    For the 12 m hit, my x-component was 10.87 m, and y-component was -5.07, as you said.

    I added the x's, and y's respectively yielding 95.64 on the x, and 39.99 on the y.
    I put that into the Pythagorean theorum, getting 103.66 m, so I added the 214 m and I get 317.66.

    The answer in the text book gives you an angle of 7.4 degrees N of E - I keep getting 22.69 degrees N of E.

    Thanks so much for taking the time to answer! I seriously appreciate it. I'm starting to understand components now :D
    Last edited: Jul 6, 2008
  5. Jul 7, 2008 #4
    OK - First thing to say is you have done well. The components you have are the same as I have .. in fact, you pretty well have all the answers you need. Also, since you have put in the effort, you get the explanation. There are places where your understanding is nearly there, but not quite.
    So try and stay with this all the way to the end, because on the way, in addition to getting the numerical answers, we want that you just do not fear vector sums of any type.

    So now, we can talk you through this, setting out what is going on. This is a problem of two separate vector additions, where the resultant of the first is used as the starting point for the second.

    The first vector we can call [tex]\,\vec{240wow}[/tex] is 240 long at an angle of 0 degrees.

    For the graphical diagram method, you can draw a vector 240mm long. (You need a BIG piece of paper). There was a time it would be very big, with very thin sharp lines, but now, we only need a visual aid, and we get to the distances using a calculator. Left to right, you are already assuming x-axis represents W-E, and y-axis represents S-N. That is OK! Probably you set its origin at the starting point.

    Now at the head of [tex]\,\vec{240wow}[/tex] you start the foot of the second vector [tex]\,\vec{96yay}[/tex].
    To help you draw it, you put a little 'x' and 'y' axes drawn in lightly with its new local origin at the start of [tex]\,\vec{96yay}[/tex].

    You know it is 96 long, and then draw it with the angle 28 degrees N of E, so rotate anticlockwise by 28 degrees.
    At this point fix in your mind that you have decided anticlockwise angles mean positive angles !

    As you have successfully calculated, the head of [tex]\,\vec{96yay}[/tex] ends up 87.76m further to the EAST, and 45.07m to the North.
    So the resultant of adding [tex]\,\vec{240wow}[/tex] to [tex]\,\vec{96yay}[/tex] is the vector from the starting point to the ending point of the second hit. He could have hit it there in the first place, but instead, it took two drives. You could have used your second equation c^2 = a^2 + b^2 -2abcosC to get the length, but we are going to duck that bit, because we already have the x-y components, so we can choose not to fall over ourselves doing what we don't need to. :smile:

    Notice that the x_component (E-W) of the first drive was 240. The y_component (N-S) was zero.
    [tex]y240=\vec{240wow}\cdot\,sin(0)\,\,=\,\,0 [/tex]

    For the second drive, you have calculated the answers - but this is how they fit.
    [tex]y96=\vec{96yay}\cdot\,sin(28)\,\,=\,\,45.07 [/tex]

    Now we can add a third vector. It can be seen as adding onto the resultant of the first two, or simply all three adding together heads-to-tails. The key thing here is vector addition is what is known as associative, which is a geek word meaning.. "you can add them up grouped any which way you like". If 3 vectors were called u, v, and w
    u + (v + w) = (u + v) + w ... easy enough.

    Now we add in the final vector, [tex]\,\vec{12duh}[/tex]

    As before, at its start point, it gets a little local x-y axes, so we can point its direction 25 degrees S of East. . The angle you need can be thought of in two ways. It was a positive angle of 335 degrees (remember going anti-clockwise), or it was -25 degrees (going clockwise). Either way will work!

    [tex]x12=\vec{12duh}\cdot\,cos(-25)\,\,=\,\,\vec{12duh}\cdot\,cos(335)\,\,=\,\,10.87[/tex] , which was an answer you got.
    [tex]y12=\vec{12duh}\cdot\,sin(-25)\,\,=\,\,\vec{12duh}\cdot\,sin(335)\,\,=\,\,-5.07 [/tex] , which by now you can see means it gets back closer to the line of the first hit.

    The last part teaches you that once all the vectors are resolved into their x and y components, you can add and subtract the components, and still end up with the right vector.
    All the x-components were found by [magnitude] x cos(angle)
    All the y-components were found by [magnitude] x sin(angle)

    So we come to the hole-in-one.
    Add up all the x-components, and get 338.64
    Add up all the y-components, and get 39.997 (OK then make it 40)

    Go on then. The grand final vector [tex]\,\vec{big.hit}[/tex] will have a magnitude that had better be (a bit) greater than 340. Pythagoras!

    If you know all sides of that big final right triangle, you can probably figure the angle in it.
    You should then have the [magnitude] and [direction] of the hole-in-one hit.
    Last edited: Jul 7, 2008
  6. Jul 7, 2008 #5
    Thanks so much! I know where my mistake was coming from (I wasn't adding the first 214 m long shot into the x-component, I was simply adding it to the end, and that's why I wasn't getting the 7.29 degrees as my direction - which, now I get problemlessly). That really clarified everything for me, though. Thanks again for taking the time! :)
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