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Math Help - Don't understand projectile motion

  1. #1
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    Don't understand projectile motion

    Hello I am working on projectile motion and I am really stuck

    I have been shown this graph:
    Don't understand projectile motion-graph_818x388.jpg

    And the parabola shape is meant to represent the trajectory and the bold line is a vector which is meant to represent initially velocity. We are ignoring wind resistance so the vertical acceleration is affected by gravity and the horizontal acceleration is 0

    I have a few questions and I really appreciate it if anyone can help me

    1. is the vector line and the 2 lines it projects onto the (horizontal and vertical) axis measured as velocity e.g. in m/s instead of m

    2. If it is measured as a velocity say m/s then wouldn’t that mean it would have multiple horizontal velocities which doesn’t make sense because it should be constant.
    For example I made up these values
    Don't understand projectile motion-modified.jpg
    As you can see there are two horizontal velocities (5 and 10) which shouldn’t be because it should be constant

    3. Why is it that the parts projected from the velocity line represent the horizontal and vertical velocity?

    Thanks
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  2. #2
    MHF Contributor Unknown008's Avatar
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    1. They are still velocities, and hence, keep m/s

    2. No, it has only one horizontal and one vertical component in one plane, because if you take the smaller ones, you would have got another resultant velocity, which is not what was given. You can 'complete the parallelogram of forces' (or rather, of velocities)to see by yourself.

    3. Have you studied how a force can be expressed into two or more components? Here, the same principle is used and to make it easier, the horizontal and vertical axes are chosen.
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    2. No, it has only one horizontal and one vertical component in one plane, because if you take the smaller ones, you would have got another resultant velocity, which is not what was given. You can 'complete the parallelogram of forces' (or rather, of velocities)to see by yourself.
    But just the fact that you can take another x value (horizontal velocity) doesn't that mean that there are other x velocities at other times, which shouldn't be possible. Can you go into a bit more depth please?


    Quote Originally Posted by Unknown008 View Post
    3. Have you studied how a force can be expressed into two or more components? Here, the same principle is used and to make it easier, the horizontal and vertical axes are chosen.
    No i haven't, what should i look up to find out more?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by deltasalt View Post
    But just the fact that you can take another x value (horizontal velocity) doesn't that mean that there are other x velocities at other times, which shouldn't be possible. Can you go into a bit more depth please?
    No, if you want it to have a second horizontal component relative to the ground, you will have to deal with 3 dimensions. It's not that it's more complicated, it's that you'll take more time to get the same result, and you get prone to make more mistakes.

    Let's in your question, the horizontal component is v \cos\theta and the vertical component is v\ sin \theta.

    If your horizontal component is equal to 10, and your vertical component equal to 11, then you get v = \sqrt{221} and \theta = 47.7^o

    Now, you take the other components, hence, v \cos\theta = 5 and v \sin \theta = 6

    This would give v = \sqrt{61} and \theta = 50.1^o

    As you can see, having different components would mean that you are throwing at different speeds v (plus as you approximated the values, you also get different angles of projection). Only a specific velocity v has a certain pair of horizontal and vertical components.

    Quote Originally Posted by deltasalt View Post
    No i haven't, what should i look up to find out more?
    You could look more into "resolving forces". This is a process where you 'break' a force into usually 2 components which are easier to deal with.

    The same principle is applied when dealing with velocity.

    You might have learned about resultant forces and resultant velocities, well, resolving forces is the opposite fo that.
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  5. #5
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    Quote Originally Posted by Unknown008 View Post
    No, if you want it to have a second horizontal component relative to the ground, you will have to deal with 3 dimensions. It's not that it's more complicated, it's that you'll take more time to get the same result, and you get prone to make more mistakes.

    Let's in your question, the horizontal component is v \cos\theta and the vertical component is v\ sin \theta.

    If your horizontal component is equal to 10, and your vertical component equal to 11, then you get v = \sqrt{221} and \theta = 47.7^o

    Now, you take the other components, hence, v \cos\theta = 5 and v \sin \theta = 6

    This would give v = \sqrt{61} and \theta = 50.1^o

    As you can see, having different components would mean that you are throwing at different speeds v (plus as you approximated the values, you also get different angles of projection). Only a specific velocity v has a certain pair of horizontal and vertical components.
    But couldn't you have a velocity of v = \sqrt{221} at one time and then later have a velocity of v = \sqrt{61}

    I may have figured out were i am getting confused. I haven't really worked with vectors before. Are the 3 lines in bold vectors?

    I have been thinking of it as a linear graph (which would have multiple x values) that shows velocity.

    But is a vector different to a linear graph in that it only has one value (its length) rather then multiple x and y values?
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  6. #6
    MHF Contributor Unknown008's Avatar
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    In this problem, no, you can never have a velocity of v = \sqrt{61} because the least velocity it can attain is when the vertical velocity is zero and only the horizontal velocity is present, that is a velocity of 10 m/s, and this occurs at the highest point reached. And at a different velocity, you will have a different angle, hence a different distance the projectile will cover. Everything changes when you change either the value of the initial velocity, its angle of projection, or the acceleration along the vertical. Hence why there is only one possible vertical and horizontal component.

    Which 3 lines are you referring to?

    You could say that... but it's more complicated. It is a position graph at multiple times.
    At time 0, the projectile is at (0, 0)
    At time 1, the projectile is at (10, 11)
    And so on.

    First, a vector consists of a magnitude (that you call 'value' here) and then it has a direction. A graph is often used to describe a vector. If you are comparing a line with a vector, like the line y = x with a vector, they are different things.
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  7. #7
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    Quote Originally Posted by deltasalt View Post
    No i haven't, what should i look up to find out more?
    recommend you start here ...

    What is a Projectile?
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  8. #8
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    Quote Originally Posted by Unknown008 View Post
    In this problem, no, you can never have a velocity of v = \sqrt{61} because the least velocity it can attain is when the vertical velocity is zero and only the horizontal velocity is present, that is a velocity of 10 m/s, and this occurs at the highest point reached. And at a different velocity, you will have a different angle, hence a different distance the projectile will cover. Everything changes when you change either the value of the initial velocity, its angle of projection, or the acceleration along the vertical. Hence why there is only one possible vertical and horizontal component.

    Which 3 lines are you referring to?

    You could say that... but it's more complicated. It is a position graph at multiple times.
    At time 0, the projectile is at (0, 0)
    At time 1, the projectile is at (10, 11)
    And so on.

    First, a vector consists of a magnitude (that you call 'value' here) and then it has a direction. A graph is often used to describe a vector. If you are comparing a line with a vector, like the line y = x with a vector, they are different things.
    Ok so I think I can see where I have gone wrong. Can you please tell me if I am right?

    So before I was getting mixed up and on this diagram
    http://img834.imageshack.us/img834/9272/picture1tr.jpg


    I thought that that the part in bold (the vector) was a linear graph (polynomial degree 1) and that each x velocity corresponded with a y velocity for different times. So i thought that that particular vector showed the vertical velocity at multiple times

    Like this:
    http://img686.imageshack.us/img686/5939/picture2jd.jpg

    But now I can see that the line isn’t a linear graph but is instead a vector and rather than vectors showing multiple times (like a v-x velocity graph) it just shows the velocity and direction for that point in time. Then another vector will show the velocity for another point in time.

    Like this:
    http://img839.imageshack.us/img839/4191/picture3im.jpg

    Is that correct? or am i still looking at it wrong?
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  9. #9
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    Yes, that's right, though your V2 isn't so convincing .

    The 'diagonal' upwards to the right should be tangent to the curve and should end where the horizontal and vertical velocity components end, like this:



    And the value of x, which is the horizontal component should be equal everywhere, for every component since there is no force acting along the horizontal (air resistance neglected)

    If you want really to have a graph, you'll see that this graph is a quadratic graph, of the form y = x^2. What it is exactly, is a description of the path of the projectile.
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  10. #10
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    Quote Originally Posted by Unknown008 View Post
    Yes, that's right, though your V2 isn't so convincing .

    The 'diagonal' upwards to the right should be tangent to the curve and should end where the horizontal and vertical velocity components end, like this:



    And the value of x, which is the horizontal component should be equal everywhere, for every component since there is no force acting along the horizontal (air resistance neglected)

    If you want really to have a graph, you'll see that this graph is a quadratic graph, of the form y = x^2. What it is exactly, is a description of the path of the projectile.
    Lol thanks in my defense I would have drawn that correctly on paper with a pen :P

    So would i be right in saying: when working with linear graphs (polynomial degree 1) we deal with multiple x values and multiple y values. But vectors are different to linear graphs because we only deal with the length of the line and the length of the projected lines? (as well as direction)

    and that a linear graph (polynomial degree 1) (y-x) represents multiple times. But a vector only represents an instant in time (but you can have a different vector for each time).
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  11. #11
    MHF Contributor Unknown008's Avatar
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    I don't think that it's wrong, so long it helps you understand the concept
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  12. #12
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    Quote Originally Posted by Unknown008 View Post
    I don't think that it's wrong, so long it helps you understand the concept
    Thanks now hopefully i can get these projectile motion questions out of the way before Christmas.
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