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Very Stuck on Deriving Projectile Motion Formulas

  1. Sep 6, 2012 #1
    Very Stuck on Deriving Projectile Motion Formulas

    Hey, I need some help :( . I've got an exam next Thursday, and I really suck at this stuff right now. So I need to get better! We are studying kinematics (1d and 2d).

    In this problem I need help with, we are given the attached diagram and are supposed to derive the following equations from it using the equations listed in part 2. While I've found three of them, I didn't really understand any of them, so will just list them all.

    1. The problem statement, all variables and given/known data
    Refer to the image for these.

    MJd7O.jpg
    Part A
    Find the time [itex]t_{H}[/itex] it takes the projectile to reach its maximum height [itex]H[/itex].
    Express [itex]t_{H}[/itex] in terms of [itex]v_{0}[/itex], [itex]\Theta[/itex], and [itex]g[/itex] (the magnitude of the acceleraion due to gravity).

    Part B
    Find [itex]t_R[/itex], the time at which the projectile hits the ground after having traveled through a horizontal distance [itex]R[/itex].
    Express the time [itex]t_R[/itex] in terms of [itex]v_{0}[/itex], [itex]\Theta[/itex], and [itex]g[/itex].

    Part C
    Find [itex]H[/itex], the maximum height attained by the projectile.
    Express your answer in terms of [itex]v_{0}[/itex], [itex]\Theta[/itex], and [itex]g[/itex].

    Part D:
    Find the total distance [itex]R[/itex] (often called the range) traveled in the [itex]x[/itex] direction; see the figure in the problem introduction.
    Express the range in terms of [itex]v_{0}[/itex], [itex]\Theta[/itex], and [itex]g[/itex].

    2. Relevant equations
    The 4 basic kinematics equations we have learned:
    [itex]v = v_{0} + at[/itex]
    [itex]x = x_{0} + v_{0}t + \frac{1}{2}at^{2}[/itex]
    [itex]v^{0}_{2} + 2a(x - x_{0})[/itex]
    [itex]\overline{v} = \frac{v + v_{0}}{2}[/itex]
    And these:
    [itex]v_{y0} = v_{0}sin \Theta [/itex]
    [itex]v_{x0} = v_{0}cos \Theta [/itex]

    3. The attempt at a solution
    Part A:
    I'm going to use the equation [itex]v = v_{0} + at[/itex]
    Steps...
    [itex]v - v_{0} = gt[/itex] (changed a to g for the gravity acting on the projectile)
    [itex]t = \frac{Δv}{g} [/itex] (changed the v's to change in v and divided g to isolate t)
    And that's it..But it's wrong! The correct answer says that it is [itex]t_{H} = \frac{v_{0}sin \Theta}{g}[/itex]
    So, how did they get that? I'm not seeing any errors in my math..what am I doing wrong?

    Part B:
    For this part I want to do the same thing, just with the x component ([itex]v_{0} cos \Theta [/itex]), but that's not right. The right answer is [itex]t_{H} = \frac{2v_{0}sin \Theta}{g}[/itex], which is double the previous answer. This makes sense, because the previous answer said how long it took to get halfway (the highest point is halfway). But it feels like I should be using the x component, not the y component. Can we use the x component, and if not, why, if so - how?

    Part C:
    For this part I use the following equation:
    [itex]x = x_{0} + v_{0}t + \frac{1}{2}at^{2}[/itex]
    Steps...
    The first thing I do is replace the x with y, since we are finding the max height ([itex]H[/itex])
    [itex]y = y_{0} + v_{y0}t + \frac{1}{2}at^{2}[/itex]
    Next I substitute the correct equation for [itex]t_{H}[/itex] in place of [itex]t[/itex], which was found in Part A.
    [itex]y = y_{0} + v_{y0} \frac{v_{0}sin \Theta}{g} + \frac{1}{2}a (\frac{v_{0}sin \Theta}{g})^{2}[/itex]
    Then simplify, and expanding [itex]v_{y0}[/itex], and moving [itex]y_{0}[/itex], also a = g.
    [itex]y - y_{0} = v_{0}sin \Theta \frac{v_{0}sin \Theta}{g} + \frac{1}{2}g (\frac{v_{0}sin \Theta}{g})^{2}[/itex]
    [itex]Δy = \frac{v_{0}^{2}sin^{2} \Theta}{g} + \frac{v_{0}^{2}sin^{2} \Theta}{2g}[/itex]

    And that last equation is the correct answer. Yay! I got 1 out of 3 so far (and understood it).

    Alright, one more,
    Part D
    For this one, I use the same equation as last time:
    [itex]x = x_{0} + v_{x0}t + \frac{1}{2}at^{2}[/itex]
    So, basically I'm doing the same thing as above except [itex]t[/itex] is now going to be the answer to Part B : [itex]t_{H} = \frac{2v_{0}sin \Theta}{g}[/itex] because we are wanting horizontal distance, not max height. We are going to use the x component now as well.
    Steps...
    [itex]x - x_{0} = (v_{0} cos \Theta)(\frac{2v_{0}sin \Theta}{g}) + \frac{1}{2}g(\frac{2v_{0}sin \Theta}{g})^{2}[/itex]
    [itex]Δx = \frac{2v_{0}^{2} cos \Theta sin \Theta}{g} + \frac{1}{2}g\frac{4v_{0}^{2}sin^{2} \Theta}{g^{2}}[/itex]
    [itex]Δx = \frac{2V_{0}^{2} cos \Theta sin \Theta}{g} + \frac{2v_{0}^{2} sin^{2} \Theta}{g}[/itex]

    [itex]Δx = \frac{2V_{0}^{2} cos \Theta sin \Theta + 2v_{0}^{2} sin^{2} \Theta}{g}[/itex]

    And that's all I got. I still don't know the final answer for this part. And I don't feel like I understand (barely) any of it. I don't know when acceleration due to gravity should be negative, or how the answers that are correct were gotten, or even where to start! I have read the 3 chapters in the textbook (Physics, by Giancoli) and took extensive notes on them, but still have no idea!! It's really annoying, confusing, and very frustrating. This post took a ton of time as well.

    Thanks for taking the time to read all this and any help would be very, very greatly appreciated,
    Josh
     
  2. jcsd
  3. Sep 6, 2012 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Gravity directly influences only the vertical component of motion. So if the object is projected vertically upwards with initial velocity v₀ , then, yes, you can say v=v₀ + at

    If the object is projected at some angle, then you must determine the vertical component (= v₀ · sinθ) of that initial velocity and then calculate the affect of g on that vertical component.

    Gravity does not affect the horizontal component of motion, so (if we ignore air resistance) we can say the object maintains a constant horizontal component to its velocity all the while that it's in flight.

    https://www.physicsforums.com/images/icons/icon2.gif [Broken] The actual velocity of the object at any point in time is the vector sum of those two components: the vertical velocity component and the horizontal velocity component.

    There are no shortcuts. https://www.physicsforums.com/images/icons/icon11.gif [Broken]
     
    Last edited by a moderator: May 6, 2017
  4. Sep 7, 2012 #3
    So, because the horizontal component never changes during the objects flight (ignoring air resistance) we do not use it for Part B. Instead we use the vertical component, because it is what decides how long it takes to hit the ground. Is that all right? I think I understand that now.

    But on Part A, I'm still not quite getting how they get their final answer. Let me try it again with the [itex]v[/itex] turned into the vertical compnent (as it should be).
    [itex]v_{y} = v_{y0} + at[/itex]
    [itex]v_{y} - v_{y0} = gt[/itex]
    [itex]\frac{v_{y} - v_{y0}}{g} = t[/itex]
    [itex]t = \frac{v_{y} - v_{0}sin \Theta}{g}[/itex]
    Oh! wait...Do they take off the [itex]v_{y}[/itex] because we are getting the time it gets to get to the peak of it's arc, at which point the velocity is 0. So the equation would turn into the following:
    [itex]t = \frac{0 - v_{0}sin \Theta}{g}[/itex]
    And then can you take off the negative sign because g is considered negative, thus leaving you with
    [itex] t = \frac{v_{0}sin \Theta}{g} [/itex]
    Which is the final, correct answer. Did I do this all right?
    Also, is g always a negative value, and can you turn [itex]v_{y}[/itex] into [itex]v sin\Theta[/itex], like you can turn [itex]v_{0y}[/itex] into [itex]v_{0}sin\Theta[/itex].

    Ok, and then for part b you double this previous answer, because you want the time of the full arc. But because the x velocity won't change until it touches the ground, you cannot use the x component to find this time, just the y.

    Part c I already had, and now Part D I still don't think I understand...Should I still be using the vertical component for this. So it would look like this:
    [itex]x - x_{0} = (v_{0} sin \Theta)(\frac{2v_{0}sin \Theta}{g}) + \frac{1}{2}g(\frac{2v_{0}sin \Theta}{g})^{2}[/itex]
    [itex]Δx = \frac{2v_{0}^{2} sin\Theta sin \Theta}{g} + \frac{1}{2}g\frac{4v_{0}^{2}sin^{2} \Theta}{g^{2}}[/itex]
    [itex]Δx = \frac{2v_{0}^{2}sin^{2} \Theta}{g} + \frac{2v_{0}^{2} sin^{2} \Theta}{g}[/itex]

    [itex]Δx = \frac{4v_{0}^{2}sin^{2} \Theta}{g}[/itex]

    Is this correct for Part D (finding the total distance traveled in the x direction)? This is the part I really didn't understand.

    Thanks for any help,
    Josh
     
  5. Sep 7, 2012 #4

    CAF123

    User Avatar
    Gold Member

    Another relation for the time of flight is [itex] t =\frac{s_x}{v_o\cos\theta}.[/itex] The one derived using vertical component of velocity will give the same [itex] t [/itex] because the horizontal component of velocity is constant.
    Yes.
    You can define the positive direction downwards.
    Or use the condition [tex] v_{oy}\hat{y} = -v_{fy}\hat{y} [/tex]
    No acceleration in the x direction therefore the [itex] \frac{1}{2}gt^2 [/itex] term disappears
     
  6. Sep 7, 2012 #5
    So since there is no acceleration, Part D is simply [itex]Δx = \frac{2v_{0}^{2}sin^{2}\Theta}{g}[/itex]?

    Arrgh..Just tried it and the homework program still says that it is wrong. What is still wrong with it?
     
  7. Sep 7, 2012 #6

    CAF123

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    Gold Member

    The total horizontal distance (range) is given by [itex] R = v_x t, [/itex] where [itex] v_x [/itex] is the horizontal component of velocity, and [itex] t [/itex] is the time of flight.
     
  8. Sep 7, 2012 #7
    So you do use the [itex]v_{0}{x} = v_{0} cos \Theta[/itex]? Which would change the answer to:
    [itex]Δx = \frac{2v_{0}^{2}cos\Theta sin\Theta}{g}[/itex]

    Again, I used my answer for Part B as t here. Should I not be using [itex]v_{x0}[/itex] and instead just using [itex]v_{x}[/itex]. I wouldn't think it matters, because in this case the horizontal velocity does not change after its initial, right? And is that above equation right? I don't want to try it again in the homework program because I am almost out of tries and want to get the question right.
     
  9. Sep 7, 2012 #8

    CAF123

    User Avatar
    Gold Member

    Consider [itex] v_{fx} = v_{ox} + at. [/itex] [itex] a [/itex] is zero so eqn reduces to [itex] v_{fx} = v_{ox} [/itex]
    The equation is now correct.
     
  10. Sep 7, 2012 #9
    First of all you are dealing with VECTORS. They have "unit" where the unit normally use eg. [itex]\hat{x}[/itex],[itex]\hat{y},[/itex][itex]\hat{z}[/itex] or any direction as [itex]\hat{r_\theta}[/itex]

    Here in your equation, all velocities must have identical unit.
    Like adding lenght in yards to lenght in metres.
    We need conversion for the addition.

    So your equation should be in this form.
    [itex]v = v_{0}\hat{r_\theta} + at\hat{y} [/itex]

    You are looking for v by adding velocity of v0 to velocity of at.
    v0 is going in direction of θ° to the horizontal
    "at" is a vector pointing downward(as in an arrow).
    Thus your addition is meaningless unless you do conversion(to its components).

    Many (maybe most) of the laws of physics involve vectors.
     
    Last edited: Sep 7, 2012
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