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 October 27th, 2019, 09:40 PM #1 Senior Member   Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus How to obtain the distance covered by a projectile if an additional acceleration is t The problem is as follows: A projectile is fired from a horizontal ground with a speed of $\vec{v}=10\hat{i}+40\hat{j}\,\frac{m}{s}$. Due to gravity and the wind, the shell experiences a constant acceleration of $\vec{a}=-5\left(\hat{i}+2\hat{j}\right)\frac{m}{s^{2}}$. Find in meters at what distance from the point the projectile was fired the shell hit the ground.$\begin{array}{ll} 1.&40\,m\\ 2.&80\,m\\ 3.&120\,m\\ 4.&100\,m\\ \end{array}$ I'm not very sure of how to use the vectors given to find the distance. Typically, I would use the equation of the position for the horizontal range would be: $x(t)=x_{o}+v_{o}\cos\omega t$ However since it mentions that there's an acceleration that is produced by the wind and gravity, how would I use this information? Would it mean that should I take the norm for that acceleration as follows? $\left \| \left \langle -5,-10 \right \rangle \right \|=\sqrt{125}=5\sqrt{5}$ and from that should I use the acceleration instead $g=9.8$, $g=5\sqrt{5}$ as in the equation from below? $y(t)=y_{o}+v_{o}\sin\omega t - \frac{1}{2}at^{2}$ Since it is fired from the ground would it mean: $y(t)= v_{o}\sin\omega t - \frac{1}{2}at^{2}$ ? The only information I could obtain was: $\sin\omega=\frac{40}{\sqrt{1700}}=\frac{4}{\sqrt{ 17}}$ But that's how far I went with this problem. Can somebody offer me some help with this?Supposedly, the answer is $80$, but I don't know what to do to get there. Last edited by skipjack; October 28th, 2019 at 02:07 AM. October 28th, 2019, 12:09 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,647 Thanks: 1476 You have two equations of motion, one in each direction. Vertically you have $y(t) = -10t^2 + 40t$ Horizontally you have $x(t) = -5t^2 + 10t$ We can solve the vertical equation for $t$ $0 = (40-10t)t$ $t = 0, 4$ Plugging this into the horizontal equation we get $x(4) = -5(4^2) + 10\cdot 4 = -40$ The problem just asks for distance so we take the absolute value of this. The distance from the firing point is $40m$. Selection (1). Thanks from topsquark and idontknow October 28th, 2019, 02:18 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,124 Thanks: 2332 The equations romsek used omitted a factor of $\frac12$. The second choice answer, 80 m, is correct. Thanks from topsquark, romsek, idontknow and 1 others October 28th, 2019, 07:39 AM   #4
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You mind giving us the title/author of the text source for these problems?
Attached Images proj_eq.png (1.7 KB, 19 views) Proj1.png (1.2 KB, 20 views)

Last edited by skeeter; October 28th, 2019 at 07:42 AM. October 28th, 2019, 08:33 AM   #5
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Quote:
 Originally Posted by Chemist116 $x(t)=x_{o}+v_{o}\cos\omega t$
Everything is pretty much answered (at least conceptually), but I want to make a comment about notation. Apparently I'm the only one that was confused by this but just in case.

The angle the projectile was fired at is, in your notation, $\displaystyle \omega$. It's typically given as $\displaystyle \theta$, but that's not important. What is important is how you group the symbols. The equation $\displaystyle x = x_0 + \cos \omega t$ is an equation that is usually seen for rotational motion. What you need to do, for clarity, is write
$\displaystyle x = x_0 + v_0 \cos( \omega) ~t$

-Dan

Last edited by skipjack; October 28th, 2019 at 04:05 PM. October 28th, 2019, 09:19 PM   #6
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Math Focus: Calculus Quote:
 Originally Posted by skipjack The equations romsek used omitted a factor of $\frac12$. The second choice answer, 80 m, is correct.
Geez Thanks you pointed out that skipjack. Just when I thought I understood what romsek did I felt something was missing. Indeed the answer obtained checks.  October 28th, 2019, 09:32 PM   #7
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Math Focus: Calculus Quote:
 Originally Posted by topsquark Everything is pretty much answered (at least conceptually), but I want to make a comment about notation. Apparently I'm the only one that was confused by this but just in case. The angle the projectile was fired at is, in your notation, $\displaystyle \omega$. It's typically given as $\displaystyle \theta$, but that's not important. What is important is how you group the symbols. The equation $\displaystyle x = x_0 + \cos \omega t$ is an equation that is usually seen for rotational motion. What you need to do, for clarity, is write $\displaystyle x = x_0 + v_0 \cos( \omega) ~t$ -Dan
Long story short. I have some sort of personal aversion or would you call it phobia? to the greek letters $\alpha \, \theta$ and $\beta$ and it dates back to my high school years. I'm not sure if it exists a word for describing that fear. Instead I prefer to use $\omega$ and $\phi$ as they bring me happy memories from my optics course back in university, so each time I have to work with trigonometric functions I tend to favor those letters over the "typical" ones.

Regarding the comment you mentioned about notation I did wrote my question in a rush and as I'm getting used to Latex I ommited the parenthesis between $\omega$ as most of the time I feel some obligation to use
Code:
\left( whatever goes here \right )
so that the brackets sandwich the character in the middle and do not generate errors.

I'm not sure if what I did caused any sort of confusion to the kind people who help me but if it does. I'm sorry!
The the sake of brevity and do not get tangled with a long line of Latex code I ommit these, but I'll try to use them more often next time.  October 28th, 2019, 09:41 PM   #8
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 Originally Posted by skeeter Helluva head wind ... You mind giving us the title/author of the text source for these problems?
How did you made that plot skeeter?. Perhaps is it from a Casio CFX-9860gc plus (my old trusty calc btw)?.

I wish I could know who on earth did made these problems but they are from a collection of problems on physics with unknown author.

I spotted some errors while solving some of the problems so it seems that whoever made these, probably didn't took into account real-life scenarios, a bit odd for a course that's based on describing nature and real world.

It might be out of the scope of this. But does it exist a better way to make illustrations and graphics to be used on physics or geometry?. Each time a drawing or a schematic is needed I have to make a drawing by hand in an illustrator software and it is a bit time consuming and perhaps does it exist a shortcut or a software which does have these figures prebuilt or something.  October 29th, 2019, 07:45 AM   #9
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Quote:
 Originally Posted by Chemist116 How did you made that plot skeeter?. Perhaps is it from a Casio CFX-9860gc plus (my old trusty calc btw)?.
screenshots from a TI-84 emulator

Quote:
 I wish I could know who on earth did made these problems but they are from a collection of problems on physics with unknown author.
Did you acquire them online? If so, have a link?

Quote:
 I spotted some errors while solving some of the problems so it seems that whoever made these, probably didn't took into account real-life scenarios, a bit odd for a course that's based on describing nature and real world.
Also note the authors use of speed vs velocity ... if given in component form, it's velocity.

Quote:
 It might be out of the scope of this. But does it exist a better way to make illustrations and graphics to be used on physics or geometry?. Each time a drawing or a schematic is needed I have to make a drawing by hand in an illustrator software and it is a bit time consuming and perhaps does it exist a shortcut or a software which does have these figures prebuilt or something. I use MS Paint alone or with one of a few graphing programs, depending on what I want to illustrate. October 29th, 2019, 11:15 AM   #10
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 Originally Posted by Chemist116 Long story short. I have some sort of personal aversion or would you call it phobia? to the greek letters $\alpha \, \theta$ and $\beta$ and it dates back to my high school years.
Once upon a time my AP Physics class had just started on rotational motion. I mentioned that the equations
$\displaystyle x = x_0 + v_0 t + \dfrac{1}{2}at^2$

and
$\displaystyle \theta = \theta _0 + \omega t + \dfrac{1}{2} \alpha t^2$

are essentially the same thing and they got all upset about the Greek variables. So the next day I wrote down something like
$\displaystyle \daleth = \daleth _0 + \beth _0 t + \dfrac{1}{2} \aleph t^2$

They were very happy to get back the Greek letters...

-Dan Tags acceleration, additional, covered, distance, obtain, projectile Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Chemist116 Physics 2 October 28th, 2019 10:14 PM Elxar Geometry 15 March 12th, 2015 12:49 PM beesee Physics 1 January 7th, 2014 12:58 PM MooseMoney Calculus 6 October 31st, 2012 03:16 PM poisonpoison Algebra 0 October 15th, 2011 02:00 PM

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