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 $t$ $0 = (4010t)t$ $t = 0, 4$ 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). 
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. 
October 28th, 2019, 07:39 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 
Helluva head wind ... You mind giving us the title/author of the text source for these problems? Last edited by skeeter; October 28th, 2019 at 07:42 AM. 
October 28th, 2019, 08:33 AM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,345 Thanks: 986 Math Focus: Wibbly wobbly timeywimey stuff.  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 
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  
October 28th, 2019, 09:32 PM  #7  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Quote:
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 ) 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  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  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. I spotted some errors while solving some of the problems so it seems that whoever made these, probably didn't took into account reallife 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  
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677  Quote:
Quote:
Quote:
Quote:
 
October 29th, 2019, 11:15 AM  #10  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,345 Thanks: 986 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle x = x_0 + v_0 t + \dfrac{1}{2}at^2$ 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  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
How can I find the distance covered by a stuntman jumping over an incline?  Chemist116  Physics  2  October 28th, 2019 10:14 PM 
Please Help Solve: Determine the Distance covered by the Object from A to B to C to D  Elxar  Geometry  15  March 12th, 2015 12:49 PM 
Trivial problem to obtain Distance from Velocity.  beesee  Physics  1  January 7th, 2014 12:58 PM 
Question: Accelarating speed and distance covered.  MooseMoney  Calculus  6  October 31st, 2012 03:16 PM 
Sorting by CoveredRest Distance and Time  poisonpoison  Algebra  0  October 15th, 2011 02:00 PM 