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Motion in two dimensions: Launching up an incline

  1. Aug 19, 2015 #1
    Motion in two dimensions.png

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

    (The following is taken from Sears and Zemansky’s University physics with modern physics, thirteenth edition by Young Freedman. Chapter 3 bridging problem: Launching up an incline at page 95.

    You fire a ball with an initial speed ##v_0## at an angle Φ above the surface of an incline, which is itself inclined at an angle θ above the horizontal (see the above image). (a) Find the distance measured along the incline, from the launch point to the point when the ball strikes the incline. (b) What angle Φ gives the maximum range, measured along the incline? Ignore air resistance.

    Identify and set up:
    1. Since there’s no air resistance, this is a problem in projectile motion. The goal is to find the point where the ball’s parabolic trajectory intersects the incline.
    2. Choose the x- and y-axes and the position of the origin. When in doubt, use the suggestions given in Problem-Solving Strategy 3.1 in Section 3.3
    3. In the projectile equations from Section 3.3, the launch angle α0 is measured from horizontal. What is this angle in terms of θ and Φ? What are the initial x- and y-components of the ball’s initial velocity?
    4. You’ll need to write and equation that relates x and y for points along the incline. What is this equation? (This takes just geometry and trigonometry, not physics.)

    Execute:
    1. Write the equations for the x-coordinate and y-coordinate of the ball as functions of time t.
    2. When the ball hits the incline, x and y are related by the equation that you found in step 4. Based on this, at what time t does the ball hit the incline?
    3. Based on your answer from Execute: step 2, at what coordinates x and y does the ball land on the incline? How far is this point from the launch point?
    4. What value of Φ gives the maximum distance from the launch point to the landing point? (Use your knowledge of calculus.)

    Evaluate:
    1. Check your answers for the case θ = 0, which corresponds to the incline being horizontal rather than tilted. (You already know the answers for this case. Do you know why?)

    (That is all from the book. From now on it will be me speaking. I am not a native English speaker and wasn’t taught math in English so I hope you will forgive me if I don’t manage to explain my thoughts too well.
    I should also mention that I study philosophy, but are trying to learn some physics by myself on the side, so there might be some holes in my knowledge.)

    From the same book I know that the answers are (a) ##R = \frac{2v_0{}^2}{g} \frac{cos(\theta + \phi)sin(\phi)}{cos^2(\theta)}## and (b) ##\phi = 45^{\circ}- \frac{\theta}{2}##

    I already got a friend to help me with (a) but I can’t seem to understand (b). I would really appreciate some help with it.

    2. Relevant equations

    ##\frac{d}{dx}sin(ax) = acos(ax)##
    ##\frac{d}{dx}cos(ax) = -asin(ax)##
    ##sin(a+b)= sin(a)cos(b)+cos(a)sin(b)##
    ##cos(a+b) = cos(a)cos(b)- sin(a)sin(b)##


    3. The attempt at a solution

    So my goal is to find the value of Φ which gives the highest value of R.

    My initial thought is to take the derivative of R, ##\frac{dR}{d\phi}##, and then say ##\frac{dR}{d\phi} = 0## to find the points where R is highest/lowest.

    First I use ##cos(a+b) = cos(a)cos(b)- sin(a)sin(b)## so I get ##R = \frac{2v_0{}^2}{g} \frac{sin(\phi)(cos(\phi)cos(\theta)- sin(\phi)sin(\theta))}{cos^2\theta}##

    Then I take the derivative using ##\frac{d}{dx}sin(ax) = acos(ax)## and ##\frac{d}{dx}cos(ax) = -asin(ax)##

    I get ##\frac{dR}{d\phi} = \frac{2v_0{}^2}{g} \frac{cos(\phi)(-sin(\phi)cos(\theta)- cos(\phi)sin(\theta))}{cos^2\theta}

    \leftrightarrow \frac{dR}{d\phi} = \frac{2v_0{}^2}{g} \frac{-cos(\phi)(sin(\phi)cos(\theta)+ cos(\phi)sin(\theta))}{cos^2\theta}##

    And at last, I use ##sin(a+b)= sin(a)cos(b)+cos(a)sin(b)##

    To get: ##\frac{dR}{d\phi} = \frac{2v_0{}^2}{g} \frac{-cos(\phi)sin(\phi+\theta)}{cos^2\theta}##

    This seems to give two solutions. Either:

    ##-cos(\phi) = 0## or ##sin(\phi+\theta) = 0##

    Those solutions gives the answers ##\phi = 90^{\circ}## and ##\phi = -\theta##

    If I substitute those values for ##\phi## in R I get:

    ##R = \frac{2v_0{}^2}{g} \frac{-sin(\theta)}{cos^2(\theta)}## and ##R = \frac{2v_0{}^2}{g} \frac{sin(\theta)cos(2\theta)}{cos^2\theta}##

    For an ordinary maximization problem, this would be the part where I test whether the values of Φ for ##\frac{dR}{d\phi} = 0## gives any maximum.

    Before doing that, there are a few implicit assumptions, which I would like to try to make explicit.

    First of, if ##v_0{}^2 = 0##, the values of Φ and θ won’t matter since R will be equal to 0. I will therefore assume that ##v_0{}^2 \neq 0## and since ##v_0{}^2 ## can’t be negative either, it means that ##\frac{2v_0{}^2}{g}## can be treated as a positive constant.

    Secondly, ##90^{\circ}>\theta >-90^{\circ}##. This is because ##cos^2(\theta) \neq 0## and because ##90^{\circ} >\phi + \theta>-90^{\circ}##. If ##\phi + \theta<-90^{\circ}## or ##90^{\circ} <\phi + \theta##, the ball will never hit the incline due to gravity and the equation will have no meaning.

    Lastly, based on this, ##cos(\theta + \phi)sin(\phi)## seems to be the only part of the equation which matters, since ##\frac{2v_0{}^2}{g}## can be treated as a positive constant as shown above and ## cos^2(\theta)## is irrelevant as long as ## cos^2(\theta) \neq 0## Therefore, as long as ##cos(\theta + \phi)sin(\phi)## has maximum value, R will have maximum value.

    Based on these assumptions, I calculated the value of R for ##\theta = -10##, ##\theta = 0## and ##\theta = 10## For each value of θ I calculated the value of R if ##\phi = 89^{\circ}##, if ##\phi = 90^{\circ}##, if ##\phi = 91^{\circ}##, if ##\phi = \theta -1##, if ##\phi = \theta ## and if ##\phi = \theta +1##.

    Based on the results, neither ##\phi = 90^{\circ}## or ##\phi = -\theta## appear to be a maximum.

    Of course, I already knew this since I know the answer is ##\phi = 45^{\circ}- \frac{\theta}{2}## but I completed these steps anyway, hoping to get some clue as to what to do next. Sadly that didn’t happen, and know I don’t really know where to go from here. Any help would be much appreciated.
     
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  3. Aug 19, 2015 #2

    andrevdh

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    What I would think you are suppose to do is resolve the gravitational acceleration, g, into its x- and y-components
    along the incline, gx, and perpendicular to the incline, gy. The motion of the projectile is
    then represented by accelerated motion along both x and y directions with the components of gravitational acceleration.
    You problem then reduces to finding the launch angle that gives the maximum x-coordinate along the incline (y = 0).
     
  4. Aug 19, 2015 #3

    DEvens

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  5. Aug 20, 2015 #4
    Thanks for the reply. I believe that is what I did in part a). I resolved g and ##v_0## into their x- and y-components along the incline and perpendicular to the incline. I got ##x = v_0cos(\phi)t-0,5gsin(\theta)t^2## and ##y = v_0sin(\phi)t-0,5gcos(\theta)t^2##
    Then I found that ##t = 0## or ##t = \frac{2v_0sin(\phi)}{gcos(\theta)}## for ##y = 0##.
    Lastly, I substituted t for ##\frac{2v_0sin(\phi)}{gcos(\theta)}## in ##x = v_0sin(\phi)t-0,5gcos(\theta)t^2## and got ##x = \frac{2v_0{}^2}{g} \frac{cos(\theta + \phi)sin(\phi)}{cos^2(\theta)}## So basically, the equation for R is the equation for x when y = 0.

    You might already be aware of that. In that case I have misunderstood you and would appreciate if you could explain what you mean I should do.

    Edit: Corrected several mistakes in the equations.
     
    Last edited: Aug 20, 2015
  6. Aug 20, 2015 #5
    Thanks for the reply. I had forgotten all about the rules for taking the derivative of multiple functions. Using the rule you linked to I got

    ##\frac{dR}{d\phi} = \frac{2v_0{}^2}{g} \frac{cos(\phi)(cos(\theta)cos(\phi)-sin(\theta)sin(\phi))+sin(\phi)(-sin(\phi)cos(\theta)-cos(\phi)sin(\theta))}{cos^2(\theta)}
    \leftrightarrow \frac{dR}{d\phi} = \frac{2v_0{}^2}{g} \frac{cos(\phi)(cos(\theta)cos(\phi)-sin(\theta)sin(\phi))-sin(\phi)(sin(\phi)cos(\theta)+cos(\phi)sin(\theta))}{cos^2(\theta)}
    \leftrightarrow \frac{dR}{d\phi} = \frac{2v_0{}^2}{g} \frac{cos(\phi)cos(\theta+\phi)-sin(\phi)sin(\theta+\phi)}{cos^2(\theta)}##

    I can then say ##\frac{dR}{d\phi} = 0
    \leftrightarrow \frac{2v_0{}^2}{g} \frac{cos(\phi)cos(\theta+\phi)-sin(\phi)sin(\theta+\phi)}{cos^2(\theta)} = 0
    \leftrightarrow cos(\phi)cos(\theta+\phi)-sin(\phi)sin(\theta+\phi) = 0## or ##\frac{2v_0{}^2}{gcos^2(\theta)} = 0##
    As I explained in my first post ##\frac{2v_0{}^2}{gcos^2(\theta)} \neq 0## so
    ##cos(\phi)cos(\theta+\phi)-sin(\phi)sin(\theta+\phi) = 0
    \leftrightarrow cos(\phi)cos(\theta+\phi) = sin(\phi)sin(\theta+\phi)##
    If I want to, I can say ##1 = tan(\phi)tan(\theta+\phi)##

    But that doesn’t seem to help me. I don’t know where to go from here.
     
  7. Aug 20, 2015 #6

    andrevdh

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    This is what I get for the first derivative?
     

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    Last edited: Aug 20, 2015
  8. Aug 20, 2015 #7
    I am confused. What is a and b, and what is it the derivative of?
     
  9. Aug 20, 2015 #8

    andrevdh

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    I am just "ignoring" the constant in the equation
     

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  10. Aug 20, 2015 #9

    haruspex

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    You can make life a bit easier by spotting that your expression for R can be written so that the trig functions of phi are all functions of the angle twice phi.
     
  11. Aug 21, 2015 #10
    I don't think I understand it. If I try to ignore the ##R = \frac{2v_0{}^2}{g cos^2(\theta)}## part I get the following:

    ##R = sin(\phi)cos(\phi+\theta)##

    I can then divide that into two functions:
    ##f(\phi) = sin(\phi)## and ##g(\phi) = cos(\phi+\theta) = cos(\phi)cos(\theta)-sin(\phi)sin(\theta)##

    I get ##\frac{d}{d\phi}f(\phi) = cos(\phi)## and ##\frac{d}{d\phi}g(\phi) = -sin(\phi)cos(\theta)-cos(\phi)sin(\theta) = -sin(\phi+\theta)##

    Then I can say ##\frac{d}{d\phi}R = g(\phi)\cdot\frac{d}{d\phi}f(\phi)+ \frac{d}{d\phi} g(\phi)\cdot f(\phi) \leftrightarrow \frac{d}{d\phi}R = cos(\phi)\cdot cos(\phi+\theta)-sin(\phi+\theta)\cdot sin(\phi)##

    As far as I can tell, this doesn’t look like your approach at all. For example, the last two lines ##a sin(\phi)cos(\phi)-b sin^2(\phi)## and ##\frac{a}{2}sin(2\phi)-b sin^2(\phi)##
    I don’t understand them.

    I would really like to know, what I am missing.
     
  12. Aug 21, 2015 #11
    Hmm… you mean writing the function for R so the functions for ##\phi## becomes functions for ##2\phi## (just making sure I understand)

    In doing so, would I be right in focusing on ##sin(\phi)cos(\phi+\theta)## and ignoring ##\frac{2v_0{}^2}{g cos^2\theta}## ?
     
  13. Aug 21, 2015 #12

    haruspex

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    Yes, when optimising a function you can always ignore constant factors. But I was referring to this equation
    You have in there a ##sin(\phi)cos(\phi)## term and a ##sin^2(\phi)## term. In each case you can usefully replace them using functions of ##2\phi##.
     
  14. Aug 28, 2015 #13
    Sorry for taking so long to respond. I am guessing that you are referring to
    ##sin(2a) = 2sin(a)cos(a)##
    and ##cos(2a) = cos^2(a)-sin^2(a) = 2cos^2(a)-1 = 1-2sin^2(a)##

    Using those I get ##R = \frac{2v_0{}^2}{g} \frac{cos(\theta + \phi)sin(\phi)}{cos^2(\theta)} \leftrightarrow R = \frac{v_0{}^2}{g} \frac{2cos(\theta + \phi)sin(\phi)}{cos^2(\theta)}##

    Then I can focus on ##2cos(\theta + \phi)sin(\phi)## So
    ##2cos(\theta + \phi)sin(\phi) = 2(sin(\phi)cos(\phi)cos(\theta)-sin^2(\phi)sin(\theta)) = 2 sin(\phi)cos(\phi)cos(\theta)-2 sin^2(\phi)sin(\theta)##

    Then by using ##sin(2a) = 2sin(a)cos(a)## I get ##sin(2\phi)cos(\theta)-2sin^2(\phi)sin(\theta)##

    Now the ##sin(\phi)cos(\phi)## has been converted to ##sin(2\phi)## but I can’t see a way to use ##cos(2a) = cos^2(a)-sin^2(a) = 2cos^2(a)-1 = 1-2sin^2(a)## to convert the ##sin^2(\phi)## into an expression with ##2\phi##
     
  15. Aug 28, 2015 #14

    haruspex

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    You can't?! Try harder.
     
  16. Sep 1, 2015 #15
    I have been pondering over this the last few days, but without much result.

    I have ##R= \frac{v_0{}^2}{g} \frac{sin(2\phi)cos(\theta)-2sin^2(\phi)sin(\theta)} {cos^2(\theta)}##

    I need to convert the ##2sin^2(\phi)## into ##cos(2\phi)## using
    ##cos(2a) = cos^2(a)-sin^2(a) = 2cos^2(a)-1 = 1-2sin^2(a)##

    I need to use the ##sin^2(\phi)## so I can focus on ##cos(2a) = cos^2(a)-sin^2(a) = 1-2sin^2(a)##
    Since there is no ##cos^2(\phi)## and I see no way to get an isolated ##cos^2(\phi)##, and since I have ##2 sin^2(\phi)## I have focused on ##cos(2a) = 1-2sin^2(a)##

    To be able to do this, I need to isolate the ##1-2sin^2(\phi)## in ##sin(2\phi)cos(\theta)-2sin^2(\phi)sin(\theta)## from the ##sin(2\phi)cos(\theta)## and the ##sin(\theta)##
    However, to do this I would need to make the ##sin(2\phi)cos(\theta)## equivalent to the ##sin(\theta)## somehow, so I can isolate those outside a pair of brackets with the ##1-2sin^2(\phi)## inside.

    I haven’t found a way to do this. I have tried to play around with the expressions and tried to introduce other expressions (more specifically, the ##cos^2(\theta)## to see if that yields any results, but I haven’t succeeded.
     
  17. Sep 1, 2015 #16

    andrevdh

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    Ignore the trig functions with the theta arguments - they are constant and will
    not influence the points (phi) where the maxima/minima are located, that is
    why I replaced them with a's and b's.They just make things seem uncessary
    complicated.
     
  18. Sep 1, 2015 #17

    haruspex

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    More specifically, using ##cos(2a) = 1-2sin^2(a)##. Just do it.
    Not sure what your difficulty is. You seem to be trying to convert both the sin2 term and the cos2 term in a single step. That is not necessary. Do them idependently. As andrevdh writes, don't get hung up on the theta terms. They're just arbitrary factors here.
     
  19. Sep 2, 2015 #18
    I think I am starting to get what is happening. Most likely I am making a very banal mistake which is making this a lot more complicated for me. I will try to explain my thinking and maybe you will be able to see where I am wrong.

    The way I am thinking, to use the ##cos(2a) = 1-2sin^2(a)## I will need to have something like ##sin(2\phi)cos(\theta)[1-2sin^2(\phi)]sin(\theta)##
    In other words, I need to isolate the ##1-2sin^2(\phi)## from the ##sin(2\phi)cos(\theta)## and the ##sin(\theta)## with brackets.
    To me it seems that this would be possible if I had something like ##sin(2\phi)cos(\theta)-2sin^2(\phi)sin(2\phi)cos(\theta)##
    Then I would be able to say ##sin(2\phi)cos(\theta)-2sin^2(\phi)sin(2\phi)cos(\theta) = sin(2\phi)cos(\theta)[1-2sin^2(\phi)] = cos(2\phi)sin(2\phi) cos(\theta)##

    The problem I am having is that, to me it (mistakenly?) seems that I shouldn’t be able to use ##cos(2a) = 1-2sin^2(a)## on ##sin(2\phi)cos(\theta)-2sin^2(\phi)sin(\theta)## because it has the form of ##a-2sin^2(\phi)b## rather than ##a-2sin^2(\phi)a##.
    In the last case, I would be able to say ## a-2sin^2(\phi)a = (1-2sin^2(\phi))a = cos(2\phi)a##.
    In the ##a-2sin^2(\phi)b## case, I don’t know how to handle the a and b when converting the ##1-2sin^2(\phi)## into ##cos(2\phi)##. I don’t know what the final product would be.
     
  20. Sep 2, 2015 #19

    haruspex

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    Just rearrange ##\cos(2a)=1-2\sin^2(a)## into the form ##\sin^2(a)=...##
     
  21. Sep 5, 2015 #20
    A very banal mistake indeed. That was slightly embarrassing.
    With that, the rest of the solution should look like this:

    ##sin(2\phi)cos(\theta)-2sin^2(\phi)sin(\theta)##
    ##=sin(2\phi)cos(\theta)-cos(2\phi)sin(\theta)-sin(\theta)##

    ##f(\phi)=sin(2\phi)cos(\theta)-cos(2\phi)sin(\theta)-sin(\theta)##
    ##\leftrightarrow f´(\phi)=2cos(2\phi)cos(\theta)-2sin(2\phi) sin(\theta)##
    ##f´(\phi)=0##
    ##\leftrightarrow 0=2cos(2\phi)cos(\theta)-2sin(2\phi) sin(\theta)##
    ##\leftrightarrow 0=cos(2\phi)cos(\theta)-sin(2\phi) sin(\theta)##
    ##\leftrightarrow 0=cos(2\phi+\theta)##
    ##\leftrightarrow 90=2\phi+\theta##
    ##\leftrightarrow \phi=45-\theta##

    Thanks a lot for the help. I learned quite a few things from this problem. Seeing as I don’t have access to other students or teachers, this was really valuable. If I run into a troublesome problem, I will probably try to do a few things different on my part, but I will definitely use this site again. I really appreciate your helping-philosophy.
     
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