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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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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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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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Quote:
 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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Quote:
 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

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