free4spirit Posted April 13, 2011 Share Posted April 13, 2011 Hi everyone Im new here...but love all the discussions going on.....I have a question for my uni assignment, that has totally stumped me....and ive never been very good at physics... So if anyone could shed any light it would be massively appreciated! Question: The arrow fired as described in part (a) reaches a maximum height of 125 m above the ground before starting to fall back downwards. When the arrow hits the ground, the archer notices the head penetrates 20.0 cm into the earth. By considering the energies involved, calculate the speed of the arrow when it hits the ground and then its deceleration as it penetrates the earth. Quote your answer to three significant figures and using scientific notation. (Assume the acceleration due to gravity is 9.81 m s− 2 and ignore any effects of air resistance.) Im not entirely sure what equation to use to figure this out....i thought maybe W = ΔE_k? but am still utterly confused.... so absolutely any help would make my day! cheers and hope you are all well Ax Link to comment Share on other sites More sharing options...
Spyman Posted April 13, 2011 Share Posted April 13, 2011 http://en.wikipedia.org/wiki/Acceleration#Uniform_acceleration Link to comment Share on other sites More sharing options...
free4spirit Posted April 13, 2011 Author Share Posted April 13, 2011 http://en.wikipedia....rm_acceleration Thanks very much should be able to figure this out now.... Ax Link to comment Share on other sites More sharing options...
imatfaal Posted April 13, 2011 Share Posted April 13, 2011 The question does state "By considering the energies involved" - now I would just use simple uniform acceleration equations of motion - but that will not answer the question as it has been asked. I think you need to consider the link between gravitational potential energy and kinetic energy - and explicitly show how the obvious equation of motion can be derived from the equality of PE and KE 1 Link to comment Share on other sites More sharing options...
free4spirit Posted April 13, 2011 Author Share Posted April 13, 2011 The question does state "By considering the energies involved" - now I would just use simple uniform acceleration equations of motion - but that will not answer the question as it has been asked. I think you need to consider the link between gravitational potential energy and kinetic energy - and explicitly show how the obvious equation of motion can be derived from the equality of PE and KE Aahh thank you for that ...im still working it out....ive been re-reading the question numerous times, but have got to the point where my brain is going arghh! lol thanks again Ax Link to comment Share on other sites More sharing options...
imatfaal Posted April 13, 2011 Share Posted April 13, 2011 As a further hint - take a look at this page on gravitational potential energy and this page on kinetic energy . You want to find a pair of simple equations for U and Ek which you can set to be equal and rearrange to give a simply equation for the velocity (actually velocity squared) in terms of the the height above the ground and g. You will then notice that this bears a striking resemblance to one of the equations of motion given earlier (if you make initial velocity zero which it is at apex) Come back if you are still struggling - but do put down where you have got to yourself 1 Link to comment Share on other sites More sharing options...
mooeypoo Posted April 13, 2011 Share Posted April 13, 2011 This is projectile motion. I doubt the exercise wants you to use energies, you should probably use the equations of motion for projectile motion, and they want the final velocity of the object. I assume you have a drawing. If not, produce one, it will help you. You should have it in your book, if not, the links above have it. If you want, you can tell us what you tried already and we can continue helping you from that point. 1 Link to comment Share on other sites More sharing options...
free4spirit Posted April 13, 2011 Author Share Posted April 13, 2011 As a further hint - take a look at this page on gravitational potential energy and this page on kinetic energy . You want to find a pair of simple equations for U and Ek which you can set to be equal and rearrange to give a simply equation for the velocity (actually velocity squared) in terms of the the height above the ground and g. You will then notice that this bears a striking resemblance to one of the equations of motion given earlier (if you make initial velocity zero which it is at apex) Come back if you are still struggling - but do put down where you have got to yourself Hmm i see....well im trying really very hard but i dont seem to be getting anywhere.....but i think i figured out an equation for it....:S mgh = 1/2mv^2.....not sure if that is correct, as m is mass.....and i do not have the mass of the arrow......or i am confusing myself? again thanks sooo much for your help! Ax oh and sorry for the wait....lappy decided to freeze! This is projectile motion. I doubt the exercise wants you to use energies, you should probably use the equations of motion for projectile motion, and they want the final velocity of the object. I assume you have a drawing. If not, produce one, it will help you. You should have it in your book, if not, the links above have it. If you want, you can tell us what you tried already and we can continue helping you from that point. Thanks for helping me out well you see we havent actually done projectile motion, and it is not in our book.... if i can figure out what equation to use to calculate the speed of the arrow as it hits the ground, and also the deceleration as it penetrates the ground, i think i may be able to manage....but at the moment i just cant quite grasp how to figure it out...... no, we do not have a drawing either.... its just the equation that confuse me! :S but thanks for the help! Ax ooohhh i may have a break through! ok so taking the mgh = 1/2mv^2 that becomes v = sqrt(2gh) ? so it would be v = sqrt(2x9.80x125) = 49.5m/s ? i hope thats right and makes sense?! thanks for the help peeps Ax Link to comment Share on other sites More sharing options...
imatfaal Posted April 13, 2011 Share Posted April 13, 2011 (edited) This is projectile motion. I doubt the exercise wants you to use energies, you should probably use the equations of motion for projectile motion, and they want the final velocity of the object. I would agree with you if the question did not say "By considering the energies involved" - I cannot believe a higher education assignment would use wording like that if it didn't mean it. (or am I putting too much faith in the question setter? ) I would always do a question like this with what I learned as the suvat equations Hmm i see....well im trying really very hard but i dont seem to be getting anywhere.....but i think i figured out an equation for it....:Smgh = 1/2mv^2.....not sure if that is correct, as m is mass.....and i do not have the mass of the arrow......or i am confusing myself? again thanks sooo much for your help! you have a mass term on each side of your equation... I hadn't read the last bit of your post - yep that's what I got. You can double check by using the equation v2=u2+2as - u is zero so you get v2 = 2as . You will notice this looks very like v2= 2gh and because g is a and h is s you realise it is the same. Edited April 13, 2011 by imatfaal 1 Link to comment Share on other sites More sharing options...
free4spirit Posted April 13, 2011 Author Share Posted April 13, 2011 I would agree with you if the question did not say "By considering the energies involved" - I cannot believe a higher education assignment would use wording like that if it didn't mean it. (or am I putting too much faith in the question setter? ) I would always do a question like this with what I learned as the suvat equations you have a mass term on each side of your equation... I know....but we havent done the suvat equations....theyre not even mentioned in the book....and yes dont worry i am aware of the mass! I think ive done it! well, the first part of the question anyway.....see what you think.... Find speed of arrow v as it hits the ground. Δh = 125m and we assume g = 9.81m s‾² So equations to use are; Ek = ½ mv² and ΔEg = mgΔh We assume there is no friction or air resistance. So, as there is no friction or air resistance, all the gravitational potential energy at the maximum height of the arrow will be converted into kinetic energy as it falls to the ground. ΔEg = Ek which becomes, ½ mv² = mgΔh Then I divide both sides of the resulting equation by m, to cancel m as we do not know the arrows mass. ½ mv² = mgΔh which then becomes, ½ v² = gΔh m m So then multiply both sides by 2; v² = 2gΔh Then take the square root of both side; v = √2gΔh Insert values, so; v = √2 x 9.81m s¯² x 125m = √2452.5m² s¯² = 49.5m s¯¹ thats what ive worked out....does that make sense now? fingers crossed! and cheers sooo much for sticking with me Ax I know....but we havent done the suvat equations....theyre not even mentioned in the book....and yes dont worry i am aware of the mass! I think ive done it! well, the first part of the question anyway.....see what you think.... Find speed of arrow v as it hits the ground. Δh = 125m and we assume g = 9.81m s‾² So equations to use are; Ek = ½ mv² and ΔEg = mgΔh We assume there is no friction or air resistance. So, as there is no friction or air resistance, all the gravitational potential energy at the maximum height of the arrow will be converted into kinetic energy as it falls to the ground. ΔEg = Ek which becomes, ½ mv² = mgΔh Then I divide both sides of the resulting equation by m, to cancel m as we do not know the arrows mass. ½ mv² = mgΔh which then becomes, ½ v² = gΔh m m So then multiply both sides by 2; v² = 2gΔh Then take the square root of both side; v = √2gΔh Insert values, so; v = √2 x 9.81m s¯² x 125m = √2452.5m² s¯² = 49.5m s¯¹ thats what ive worked out....does that make sense now? fingers crossed! and cheers sooo much for sticking with me Ax sorry about the two m's there...i think you know what i mean! i just copy and pasted from star office.... Ax Link to comment Share on other sites More sharing options...
mooeypoo Posted April 13, 2011 Share Posted April 13, 2011 I would agree with you if the question did not say "By considering the energies involved" - I cannot believe a higher education assignment would use wording like that if it didn't mean it. (or am I putting too much faith in the question setter? ) You're not putting too much faith in the question setter, I misread. I apologize. You're absolutely right. The question requests energy considerations, then that's what the exercise should use to solve it. Sorry, I misread. You're right! All the more emphasizing the idea of *reading questions carefully* heh.. sorry. ~mooey P.S: I would *still* draw a picture on physics questions. If you do that, you can see where you have kinetic energy and no potential vs. where you have potential and no kinetic, etc. It's a personal choice, of course, to draw a picture, but in my experience, it helps a lot more than people think initially. 1 Link to comment Share on other sites More sharing options...
free4spirit Posted April 13, 2011 Author Share Posted April 13, 2011 You're not putting too much faith in the question setter, I misread. I apologize. You're absolutely right. The question requests energy considerations, then that's what the exercise should use to solve it. Sorry, I misread. You're right! All the more emphasizing the idea of *reading questions carefully* heh.. sorry. ~mooey Haha! no no dont worry! I literally just found the equation in the back of my notebook!! yeah true....but im pretty damn sure these questions are meant to trick you....well in my last assignment i got tricked for sure! but....i am still trying to figure out an equation to calculate the deceleration as the arrow hits the ground...... maybe i'll find another equation in another notebook! i might go check... thank thank you so much for your help! Annax oh and i am aware that deceleraiton is initial velocity - final speed/ total time taken...... but i do not know what its initial velocity is..... Ax Link to comment Share on other sites More sharing options...
imatfaal Posted April 13, 2011 Share Posted April 13, 2011 oh and i am aware that deceleraiton is initial velocity - final speed/ total time taken...... but i do not know what its initial velocity is..... Ax Hey - we're getting there. v=u+at BUT, and it's a big but, you don't know t ! So we need a different equation. - you know initial velocity (you just spent last hour calculating it!!) and you know final velocity (it's stopped) so you have v and u. What else were you told? There was a new piece of information so far unused that will allow you to use the equations of motion to calculate the deceleration. 2 Link to comment Share on other sites More sharing options...
free4spirit Posted April 13, 2011 Author Share Posted April 13, 2011 Hey - we're getting there. v=u+at BUT, and it's a big but, you don't know t ! So we need a different equation. - you know initial velocity (you just spent last hour calculating it!!) and you know final velocity (it's stopped) so you have v and u. What else were you told? There was a new piece of information so far unused that will allow you to use the equations of motion to calculate the deceleration. Haha, indeed we are! Thanks to you guys! thats precisely what i have been pondering the last wee whiley.....havent been told anything else.....just the acceleration due to gravity, the max height of the arrow, the depth at which the arrow penetrates the ground and now i know the final velocity.....i have no other information....not that i can find! am at the brink of insanity!! XD lol Ax Link to comment Share on other sites More sharing options...
imatfaal Posted April 13, 2011 Share Posted April 13, 2011 Haha, indeed we are! Thanks to you guys! thats precisely what i have been pondering the last wee whiley.....havent been told anything else.....just the acceleration due to gravity, the max height of the arrow, the depth at which the arrow penetrates the ground and now i know the final velocity.....i have no other information....not that i can find! am at the brink of insanity!! XD lol Ax Be more precise. You have two problems. The first is solved - it provided the velocity when the arrow HIT the ground. Now, WRT problem 2, is that velocity the final velocity?? I was explicit in my previous post about velocities for second problem - re-read that post. You do have one more piece of information that is staring you in the face. Link to comment Share on other sites More sharing options...
swansont Posted April 13, 2011 Share Posted April 13, 2011 Then I divide both sides of the resulting equation by m, to cancel m as we do not know the arrows mass. Another way of looking at this is that the answer is independent of the mass. 1 Link to comment Share on other sites More sharing options...
roosterJcogburn Posted April 13, 2011 Share Posted April 13, 2011 We cant tell you exacty how to do it 1 Make some assumptions 2 Draw a picture 3 Decellaration is 32ft/s>2 4 stop -start-top of trajectory some equtions Regards Keith 1 Link to comment Share on other sites More sharing options...
free4spirit Posted April 13, 2011 Author Share Posted April 13, 2011 Be more precise. You have two problems. The first is solved - it provided the velocity when the arrow HIT the ground. Now, WRT problem 2, is that velocity the final velocity?? I was explicit in my previous post about velocities for second problem - re-read that post. You do have one more piece of information that is staring you in the face. yeah i know its staring me in the face......ive been at this for 4hrs now and my head is mush......so i apologise if im missing something...... but anyways i think i may have got it! well.....again see what you think Deceleration = initial velocity – final speed/ total time taken - v = u + at But we do not know t. So we know u = 49.5m s¯¹ and the final velocity v = (49.5m s¯¹ + 0) = 24.25m s¯¹ 2 So, since the mean velocity is equal to the displacement divided by time vave = D t Therefore, the time for the arrow to come to rest is t = D = 0.20m = 0.00825 s vave 24.25m s‾¹ Then lastly, for the accelerating arrow vf = at + vo So, a = (vf – vo) = (0 – 24.25m s‾¹) = -2940m s² = -2.94m s² x 10³ t 0.00825s hope it makes sense..... Ax We cant tell you exacty how to do it 1 Make some assumptions 2 Draw a picture 3 Decellaration is 32ft/s>2 4 stop -start-top of trajectory some equtions Regards Keith Thanks for your help its much appreciated my head just mush now and ive still got more questions to get through...so yeahh apologies if im being slow Ax Link to comment Share on other sites More sharing options...
swansont Posted April 13, 2011 Share Posted April 13, 2011 Since the question wants you to solve in terms of energy, perhaps you should look at kinetic energy and work. Link to comment Share on other sites More sharing options...
imatfaal Posted April 14, 2011 Share Posted April 14, 2011 Free4 - Swansont is of course right - you are still ignoring the energy. To be a little more explicit try searching on work-energy principle Link to comment Share on other sites More sharing options...
mathsgirl Posted April 14, 2011 Share Posted April 14, 2011 Free4 - Swansont is of course right - you are still ignoring the energy. To be a little more explicit try searching on work-energy principle Hi there, I am also trying to answer the last section of this question, and I still can't, even after all your helpful suggestions! I have absolutely no, no, no clue! Please can you help? I'm racking my brains and theres nothing there! I only need help with this section: "calculate its deceleration as it penetrates the earth" thanks so much! Link to comment Share on other sites More sharing options...
mooeypoo Posted April 14, 2011 Share Posted April 14, 2011 Let's start over with the question, mathsgirl: Question: The arrow fired as described in part (a) reaches a maximum height of 125 m above the ground before starting to fall back downwards. When the arrow hits the ground, the archer notices the head penetrates 20.0 cm into the earth. By considering the energies involved, calculate the speed of the arrow when it hits the ground and then its deceleration as it penetrates the earth. Quote your answer to three significant figures and using scientific notation. (Assume the acceleration due to gravity is 9.81 m s− 2 and ignore any effects of air resistance.) I suggest you draw it, but even if you don't, think about the energies you need to consider: what happens to the arrow as it reaches the top height? What happens to the arrow's energies when it reaches the floor (just before it penetrates the ground)? Can you figure out the velocity and acceleration of the arrow JUST before it hits the ground? You're told it penetrates 20.0 cm into the ground. Can you think of a way to connect the distance the arrow travels to the acceleration/deceleration that acts on it while it moves through the ground? Think about all these things, and try to see if you can start the question. If you're still stuck, show us the work you've done and we'll go from there. Link to comment Share on other sites More sharing options...
mathsgirl Posted April 14, 2011 Share Posted April 14, 2011 Let's start over with the question, mathsgirl: I suggest you draw it, but even if you don't, think about the energies you need to consider: what happens to the arrow as it reaches the top height? What happens to the arrow's energies when it reaches the floor (just before it penetrates the ground)? Can you figure out the velocity and acceleration of the arrow JUST before it hits the ground? You're told it penetrates 20.0 cm into the ground. Can you think of a way to connect the distance the arrow travels to the acceleration/deceleration that acts on it while it moves through the ground? Think about all these things, and try to see if you can start the question. If you're still stuck, show us the work you've done and we'll go from there. Thanks for this. So, when the arrow reaches its top height, the kinetic energy has been turned into gravitational potential energy and there is a pause before it turns back to kinetic energy. The velocity of the arrow JUST before it hits the ground is 49.5 ms-1. When it penetrates the ground, the kinetic energy is turned into sound energy and heat. But I still don't know how to work out the deceleration. My poor brain hurts sso much! any help appreciated! Link to comment Share on other sites More sharing options...
mooeypoo Posted April 15, 2011 Share Posted April 15, 2011 Thanks for this. So, when the arrow reaches its top height, the kinetic energy has been turned into gravitational potential energy and there is a pause before it turns back to kinetic energy. The velocity of the arrow JUST before it hits the ground is 49.5 ms-1. When it penetrates the ground, the kinetic energy is turned into sound energy and heat. But I still don't know how to work out the deceleration. My poor brain hurts sso much! any help appreciated! Okay, my first inclination is to do this part with forces rather than energies, but we have to go by your question. Here's what I would do: Alright, so now you have "Stage II" of the movement. Do the same: What energies does the arrow have the moment it hits the ground? Then, what energies does it have when it stopped? It's the same principles... (hint: Think of the ground as your origin, so y-zero/x-zero, so at some point you will start looking at the arrow traveling on the "negative" values but it STILL has "distance" away from the zero, which meaaaaans............ When you calculate "mgh", you already decide that your acceleration is "g(=9.8)". What do you do if it isn't? ~mooey Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted April 15, 2011 Share Posted April 15, 2011 Since you know the velocity when it hits the ground, you can find the kinetic energy, right? So there's a more intuitive way to work this out. As you said before, [math]W=\Delta E_k[/math]. If the arrow goes to a standstill, the final [math]E_k =0[/math], so you can figure out what [math]\Delta E_k[/math] must be. Then you know that [math]W = Fd[/math] -- that is, work is force times distance. And since [math]F=ma[/math], [math]W = mad[/math]. You know the distance the arrow travels in the ground and its mass. Run with it. Link to comment Share on other sites More sharing options...
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