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October 27th, 2019, 09:40 PM   #1
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Question 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.
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October 28th, 2019, 12:09 AM   #2
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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).
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October 28th, 2019, 02:18 AM   #3
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The equations romsek used omitted a factor of $\frac12$.

The second choice answer, 80 m, is correct.
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October 28th, 2019, 07:39 AM   #4
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Helluva head wind ...

You mind giving us the title/author of the text source for these problems?
Attached Images
File Type: png proj_eq.png (1.7 KB, 19 views)
File Type: png Proj1.png (1.2 KB, 20 views)
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Last edited by skeeter; October 28th, 2019 at 07:42 AM.
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October 28th, 2019, 08:33 AM   #5
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Quote:
Originally Posted by Chemist116 View Post
$x(t)=x_{o}+v_{o}\cos\omega t$
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
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Last edited by skipjack; October 28th, 2019 at 04:05 PM.
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October 28th, 2019, 09:19 PM   #6
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Quote:
Originally Posted by skipjack View Post
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.
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October 28th, 2019, 09:32 PM   #7
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Quote:
Originally Posted by topsquark View Post
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.
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October 28th, 2019, 09:41 PM   #8
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Quote:
Originally Posted by skeeter View Post
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.
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October 29th, 2019, 07:45 AM   #9
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Quote:
Originally Posted by Chemist116 View Post
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.
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October 29th, 2019, 11:15 AM   #10
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Quote:
Originally Posted by Chemist116 View Post
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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