downcast Posted December 6, 2011 Share Posted December 6, 2011 A ball is thrown upward with a velocity of 39.2 m/s, Four seconds later, a second ball is thrown up with the same velocity. At what height will the two balls meet? [ can someone give me even just an idea on how to solve this ? haaa . im really having a hard time on this one .. thank you ! ] Link to comment Share on other sites More sharing options...

Daedalus Posted December 6, 2011 Share Posted December 6, 2011 (edited) This one is actually quite easy, but I can't do your homework for you : ) Can you show us what you have tried so far and the results you have obtained? I or anyone else can then help guide you to the correct answer. I can at least give you a few hints in that you need the equation which defines position in relation to acceleration, initial velocity, and initial position as follows: [math]y=\frac{1}{2}\, a \, t^2 + v_0 \, t + y_0[/math] You should know the acceleration provided by gravity, [math]-9.8 \ m / s^2[/math]. You also stated the initial velocity, [math]39.2 \ m / s[/math]. Since you didn't specify an initial position, we can safely assume that both balls are thrown from the same height and you can zero out the [math]y_0[/math] variable and simplify the equation to: [math]y=\frac{1}{2}\, \left(-9.8 \ m / s^2\right) \, t^2 + \left(39.2 \ m / s\right) \, t + \left(0 \ m\right)=\left(-4.9 \ m / s^2\right) \, t^2 + \left(39.2 \ m / s\right) \, t[/math] Next, you need to know how to set up the problem. The key here is that both balls travel the same path except one ball is thrown four seconds later. This involves modifying the equation to account for the offset in time of four seconds for the second ball. Then, set the equations for both balls equal to each other to solve for the time their positions are equal. I'll be happy to help you out once I have seen where you are getting stuck. Edited December 6, 2011 by Daedalus Link to comment Share on other sites More sharing options...

csmyth3025 Posted December 6, 2011 Share Posted December 6, 2011 This one is actually quite easy, but I can't do your homework for you : ) Can you show us what you have tried so far and the results you have obtained? I or anyone else can then help guide you to the correct answer. I can at least give you a few hints in that you need the equation which defines position in relation to acceleration, initial velocity, and initial position as follows: [math]y=\frac{1}{2}\, a \, t^2 + v_0 \, t + y_0[/math] You should know the acceleration provided by gravity, [math]-9.8 \ m / s^2[/math]. You also stated the initial velocity, [math]39.2 \ m / s[/math]. Since you didn't specify an initial position, we can safely assume that both balls are thrown from the same height and you can zero out the [math]y_0[/math] variable and simplify the equation to: [math]y=\frac{1}{2}\, \left(-9.8 \ m / s^2\right) \, t^2 + \left(39.2 \ m / s\right) \, t + \left(0 \ m\right)=\left(-4.9 \ m / s^2\right) \, t^2 + \left(39.2 \ m / s\right) \, t[/math] Next, you need to know how to set up the problem. The key here is that both balls travel the same path except one ball is thrown four seconds later. This involves modifying the equation to account for the offset in time of four seconds for the second ball. Then, set the equations for both balls equal to each other to solve for the time their positions are equal. I'll be happy to help you out once I have seen where you are getting stuck. I don't know with what level of math you're familiar. The equations and reasoning given by Daedalus should be helpful if your question is typical of the math you're doing now. As a general observation, I'm sure that you realize the the first ball will always be ahead of the second ball on the way up. Only after the first ball has reached the apex of its trajectory and is on the way down will both balls meet (while the second ball is, of course, still going up). The formula given should point you in the right direction to find out how many seconds will transpire until that meeting occurs. Chris Link to comment Share on other sites More sharing options...

Daedalus Posted December 7, 2011 Share Posted December 7, 2011 (edited) Sometimes it is helpful to visualize the problem before attempting to solve it. I have created the following graph so that you can see the mechanics of both balls in relation to height and time: Once you have solved for the time that the two balls will meet, you can plug that value into one of the equations to obtain the answer you need. Also, it is always a good habit to validate your answer by checking both equations to see if they produce the same results. I have purposely erased the units as to not give away the answer. Edited December 7, 2011 by Daedalus Link to comment Share on other sites More sharing options...

Schrödinger's hat Posted December 7, 2011 Share Posted December 7, 2011 Psst he already got the answer in chat. Although bailed before I could show him the graph. 1 Link to comment Share on other sites More sharing options...

Daedalus Posted December 7, 2011 Share Posted December 7, 2011 (edited) Psst he already got the answer in chat. Although bailed before I could show him the graph. Awe... I see. Well then, I guess it won't hurt to show everyone how to find the solution for the sake of anyone else who needs help solving this type of problem : ) As stated before, the equation for the path of a body considering uniform acceleration, initial velocity and initial position is as follows: [math]y=\frac{1}{2}\, a \, t^2 + v_0 \, t + y_0[/math] The acceleration provided by gravity is [math]-9.8 \ m / s^2[/math], the stated initial velocity is [math]39.2 \ m / s[/math], and the second ball is thrown [math]4 \, s[/math] after the first ball. Since the problem didn't specify an initial position, we can safely assume that both balls are thrown from the same height and you can zero out the [math]y_0[/math] variable giving us the equation for the first ball as follows: [math]y=\left(-4.9 \ m/s^2\right) \, t^2 + \left(39.2 \ m / s\right) \, t[/math] The second ball is thrown four seconds later. The means that we must modify the equation for the second ball such that by the time four seconds have past, the second ball has started following the path of the first ball. This means the we must subtract four seconds from the time variable in the equation for the second ball. This may seem counterintuitive. But we can see that by subtracting four seconds from the time variable for the second ball, we will have placed the second ball at the starting position of the first ball: The first ball has been in flight for four seconds. The second ball starts at the same position as the first ball four second later. Such that [math]t - 4\, s[/math] will offset the time for the second ball and place it at the position the first ball was located at (i.e. Starting time for ball2 in relation to ball1 is [math]4\, s-4\, s = 0 \, s[/math]). This gives us the equation for the path of the second ball in relation to the first ball: [math]y=\left(-4.9 \ m/s^2\right) \, (t - 4 \, s)^2 + \left(39.2 \ m / s\right) \, (t - 4 \, s)[/math] The problem states that we need to find the position where both balls meet. Because we now know the equations for the path of both balls, all we have to do is determine where both paths return the same height. This is solved by setting the equations equal to each other which yields: [math]\left(-4.9 \ m/s^2\right) \, t^2 + \left(39.2 \ m / s\right) \, t = \left(-4.9 \ m/s^2\right) \, (t - 4 \, s)^2 + \left(39.2 \ m / s\right) \, (t - 4 \, s)[/math] The first thing we need to do is expand the right hand side. I am going to assume the reader knows how to expand the term, [math]\left(-4.9 \ m/s^2\right) \, (t - 4 \, s)^2[/math], using the FOIL method and how to distribute the specified constants. [math]\left(-4.9 \ m/s^2\right) \, t^2 + \left(39.2 \ m / s\right) \, t = \left(-4.9 \ m/s^2\right) \, t^2 + \left(78.4 \ m / s\right) \, t - \left(235.2 \, m\right)[/math] Next, we will add [math]\left(4.9 \ m/s^2\right) \, t^2[/math] to both sides: [math]\left(39.2 \ m / s\right) \, t = \left(78.4 \ m / s\right) \, t - \left(235.2 \, m\right)[/math] Then, we will subtract [math]\left(78.4 \ m / s\right) \, t[/math] from both sides: [math]\left(-39.2 \ m / s\right) \, t = \left(-235.2 \, m\right)[/math] Finally, we will divide both sides by [math]\left(-39.2 \ m / s\right)[/math]: [math]t = \frac{-235.2 \, m}{-39.2 \ m / s}=6.0 \, s[/math] Now that we know the time that both balls will meet, we can plug this result back into our equation for the first ball and solve for the height that the collision took place: [math]y=\left(-4.9 \ m/s^2\right) \, (6.0 \, s)^2 + \left(39.2 \ m / s\right) \, (6.0 \, s)=58.8\, m[/math] Check the answer with the equation for the path of the second ball: [math]y=\left(-4.9 \ m/s^2\right) \, (6.0 \, s - 4 \, s)^2 + \left(39.2 \ m / s\right) \, (6.0 \, s - 4 \, s)=58.8\, m[/math] We can see that this is the correct solution and further validate our result by graphing it out: We can also generalize the process for solving this type of problem. Because gravity will always be [math]-9.8\ m/s^2[/math] when solving these type of basic physics problems, we do not have to worry about using the quadratic formula to find both solutions because the term, [math]\left(-9.8 \ m/s^2\right)\, t^2[/math], will always cancel out. This allows us to yield the following formula for these types of problems: The equation for the path of the first body: [math]y=\frac{1}{2}\, a \, (t-t_1)^2 + v_1 \, (t-t_1) + y_1[/math] The equation for the path of the second body: [math]y=\frac{1}{2}\, a \, (t-t_2)^2 + v_2 \, (t-t_2) + y_2[/math] Such that: [math]\frac{1}{2}\, a \, (t-t_1)^2 + v_1 \, (t-t_1) + y_1 = \frac{1}{2}\, a \, (t-t_2)^2 + v_2 \, (t-t_2) + y_2[/math] Yields the following generalization: [math]t=\frac{a\, \left(t_1^2 - t_2^2\right)-2\, \left(t_1 \, v_1 - t_2\, v_2\right)+2\, \left(y_1-y_2\right)}{2\, a\, \left(t_1 - t_2\right) -2\, \left( v_1 - v_2\right)} [/math] Edited December 7, 2011 by Daedalus 1 Link to comment Share on other sites More sharing options...

downcast Posted December 8, 2011 Author Share Posted December 8, 2011 Got the same answer. Thanks to Schrödinger =) Link to comment Share on other sites More sharing options...

csmyth3025 Posted December 9, 2011 Share Posted December 9, 2011 First, thanks Daedalus for your thorough and very instructive explanation of this problem. I have a very rudimentary idea of how you're able to display formulae using LaTeX code thanks to the LaTeX tutorial thoughtfully pinned at the beginning of the Mathematics section of this forum . For instance: [latex]c^2=a^2+b^2[/latex]. What feature or code produces the excellent graph you were able to include in your post? Chris As an afterthought, I thought I'd try using ["math"][/"math"] encoding instead of ["latex"][/"latex"] coding. I get: [math]c^2=a^2+b^2[/math] and: [latex]c^2=a^2+b^2[/latex] in both cases. I suppose it's a matter of personal preference which tag one uses in this forum. Chris Link to comment Share on other sites More sharing options...

Schrödinger's hat Posted December 9, 2011 Share Posted December 9, 2011 (edited) First, thanks Daedalus for your thorough and very instructive explanation of this problem. I have a very rudimentary idea of how you're able to display formulae using LaTeX code thanks to the LaTeX tutorial thoughtfully pinned at the beginning of the Mathematics section of this forum . For instance: [latex]c^2=a^2+b^2[/latex]. What feature or code produces the excellent graph you were able to include in your post? This is just an image. You can include images from a url or as attachments. I don't recognise the software Daedalus used for his graph, but one free package is octave, it's an open source version of matlab using the gnuplot package for graphics (you can also install other graphing packages). Graphics can be produced directly in latex. [latex] \setlength{\unitlength}{1mm} \begin{picture}(60, 40) \put(20,30){\circle{1}} \put(20,30){\circle{2}} \put(20,30){\circle{4}} \put(20,30){\circle{8}} \put(20,30){\circle{16}} \put(20,30){\circle{32}} \put(40,30){\circle{1}} \put(40,30){\circle{2}} \put(40,30){\circle{3}} \put(40,30){\circle{4}} \put(40,30){\circle{5}} \put(40,30){\circle{6}} \put(40,30){\circle{7}} \put(40,30){\circle{8}} \put(40,30){\circle{9}} \put(40,30){\circle{10}} \put(40,30){\circle{11}} \put(40,30){\circle{12}} \put(40,30){\circle{13}} \put(40,30){\circle{14}} \put(15,10){\circle*{1}} \put(20,10){\circle*{2}} \put(25,10){\circle*{3}} \put(30,10){\circle*{4}} \put(35,10){\circle*{5}} \end{picture} [/latex] (copied and pasted from http://en.wikibooks....eating_Graphics ) But plotting directly is cumbersome without extra packages which this forum doesn't seem to have installed. The forum also tends to strip out carriage returns from latex for some reason. This makes things a bit hard to read at times. Edited December 9, 2011 by Schrödinger's hat Link to comment Share on other sites More sharing options...

Daedalus Posted December 9, 2011 Share Posted December 9, 2011 (edited) First, thanks Daedalus for your thorough and very instructive explanation of this problem. I have a very rudimentary idea of how you're able to display formulae using LaTeX code thanks to the LaTeX tutorial thoughtfully pinned at the beginning of the Mathematics section of this forum . For instance: [latex]c^2=a^2+b^2[/latex]. What feature or code produces the excellent graph you were able to include in your post? Chris As an afterthought, I thought I'd try using ["math"][/"math"] encoding instead of ["latex"][/"latex"] coding. I get: [math]c^2=a^2+b^2[/math] and: [latex]c^2=a^2+b^2[/latex] in both cases. I suppose it's a matter of personal preference which tag one uses in this forum. Chris As Schrodinger's hat stated, it is an image that I created by plotting the equations in Mathematica. I then attached the image to the post. I have Mathematica version 5.2 but the newer version includes anti-aliasing which produces smoother looking curves. Edited December 10, 2011 by Daedalus Link to comment Share on other sites More sharing options...

csmyth3025 Posted December 11, 2011 Share Posted December 11, 2011 As Schrodinger's hat stated, it is an image that I created by plotting the equations in Mathematica. I then attached the image to the post. I have Mathematica version 5.2 but the newer version includes anti-aliasing which produces smoother looking curves. Thanks. I'm ponderously working my way through an introduction to calculus textbook (it's hard to teach old dogs new tricks). I know I can graph equations like this on my T-83 calculator. Once I get the hang of it I would like to give Mathematica a try. This thread has been very helpful to me. Chris Link to comment Share on other sites More sharing options...

Daedalus Posted December 11, 2011 Share Posted December 11, 2011 (edited) I'm glad we can be of some help : ) Edited December 11, 2011 by Daedalus Link to comment Share on other sites More sharing options...

Schrödinger's hat Posted December 12, 2011 Share Posted December 12, 2011 Thanks. I'm ponderously working my way through an introduction to calculus textbook (it's hard to teach old dogs new tricks). I know I can graph equations like this on my T-83 calculator. Once I get the hang of it I would like to give Mathematica a try. This thread has been very helpful to me. Chris Mathematica has a free fifteen day trial, but for more than that you'll have to pay. Some free alternatives: Sage. Online notebook or download it Octave Maxima Of these, Sage is probably the closest to Mathematica in capabilities (although last time I used it many of the convenience features like automatic inclusion of constants and plain text input were still missing. Maxima is closest the Mathematica in syntax/style. All of Maxima's features are available through Sage (sage started as a front end for a number of different projects, although they have more of their own back-end now). All of these (and mathematica) are very functional in their philosophy. ie. define this function, plot it over this range, differentiate it, and so on. (This doesn't mean they can't handle lists of numbers though) Octave is much more like Matlab (there's a lot of code/input that will work directly on both). Its features are also available through Sage (as an optional install last time I looked). Octave and Matlab are both very imperative in style. Define this list of numbers, apply this operation to all of them, find the difference between each and divide it by a constant, plot the resulting list of numbers and so on. They have a variety of ways of doing symbolic stuff, but it's primarily numeric. Being a super-set of many other packages, Sage is probably the most capable. I have also heard good things about Goegebra I haven't used it, but the philosophy seems to be more about presenting things visually and doing geometry. R is another one. The main benefit is that it has been around (and open source) for a very long time. As such you're likely to find libraries/source to do whatever. I've only really used it in the context of statistics. It's very good for analyzing/plotting/doing stats on lots of data. I don't know about modelling/symbolic stuff as I've never used it for such. Also Capnrefsmmat responded favourably to my suggestion of adding some pstricks libraries to the latex installation. When he's done with exams I anticipate you shall be able to use latex to plot stuff directly on the forums. 1 Link to comment Share on other sites More sharing options...

Daedalus Posted December 12, 2011 Share Posted December 12, 2011 (edited) Schrödinger's hat has definitely listed some great alternatives to Mathematica. You can also use the following link to view most of these programs along with licensing costs. I'm not sure how up-to-date the pricing is, but it should give you a general idea : ) http://en.wikipedia....algebra_systems Also Capnrefsmmat responded favourably to my suggestion of adding some pstricks libraries to the latex installation. When he's done with exams I anticipate you shall be able to use latex to plot stuff directly on the forums. That would be awesome!!! Edited December 12, 2011 by Daedalus Link to comment Share on other sites More sharing options...

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