Hi,
I'm really confused by this question. Say you have a ball at point P that is going to be projected at 30 degrees to the horizontal, with a velocity of Ums1. You have a goal 25 metres away from P, and a goal keeper 10 metres infront of P between the goal and P. Like this...
P<====10m====>goalkeeper<======1 5m=====> Goal
The ball will be scored as long as the ball is more than 3 metres above the ground when it reaches the goal keeper but less than 3 metres above the ground when it reaches the goal. What is the range of values of U for which the ball will be scored?
I'm utterly flummoxed by this question.
I'd really appreciate any help.
A difficult M2 question (difficult for me, may not be difficult for you)
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 27072016 18:42
Last edited by student1856; 27072016 at 18:44. 
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 27072016 19:01
(Original post by student1856)
Hi,
I'm really confused by this question. Say you have a ball at point P that is going to be projected at 30 degrees to the horizontal, with a velocity of Ums1. You have a goal 25 metres away from P, and a goal keeper 10 metres infront of P between the goal and P. Like this...
P<====10m====>goalkeeper<======1 5m=====> Goal
The ball will be scored as long as the ball is more than 3 metres above the ground when it reaches the goal keeper but less than 3 metres above the ground when it reaches the goal. What is the range of values of U for which the ball will be scored?
I'm utterly flummoxed by this question.
I'd really appreciate any help.
As the horizontal speed is assumed to be constant, you can find the time that it would take to reach the goalkeeper and the goal, in terms of u. If we split this up into 2 scenarios, where scenario 1 is when it needs to be above the goalkeeper, and the scenario 2 is in the goal.
Scenario 1, the vertical displacement must be greater than 3, we know the time this occurs, we know the initial speed of the ball in the vertical direction which is , we also know the acceleration due to gravity. Using the suvat equation we can solve for u. Same approach for the second scenario where the vertical displacement must be less than 3. You should get two inequalities which squeeze the values of u. I'm going to try it now too .
Edit: for the times, the goalkeeper is 10m away, hence hence Now the goal is 25m away, so swap 25 for 10 in the equation, giving you the time it takes to reach there.Hope this helps!Last edited by AMarques; 27072016 at 19:51.Post rating:1 
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 27072016 19:05
(Original post by AMarques)
Hey
As the horizontal speed is assumed to be constant, you can find the time that it would take to reach the goalkeeper and the goal, in terms of u. If we split this up into 2 scenarios, where scenario 1 is when it needs to be above the goalkeeper, and the scenario 2 is in the goal.
Scenario 1, the vertical displacement must be greater than 3, we know the time this occurs, we know the initial speed of the ball in the vertical direction which is , we also know the acceleration due to gravity. Using the suvat equation we can solve for u. Same approach for the second scenario where the vertical displacement must be less than 3. You should get two inequalities which squeeze the values of u. I'm going to try it now too Edit: for the times, the goalkeeper is 10m away, hence hence Now the goal is 25m away, so swap that instead of 10, giving you the time it takes to reach there.Hope this helps!
I understand what you're saying and it makes sense. However, you said that you have the time that this occurs. How do you know t? I see your edit, I'll also attempt it now.Last edited by student1856; 27072016 at 19:06. 
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 27072016 19:13
(Original post by student1856)
Hi,
I understand what you're saying and it makes sense. However, you said that you have the time that this occurs. How do you know t? I see your edit, I'll also attempt it now. 
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 27072016 19:16
(Original post by AMarques)
My final answers to 1dp were 15.3 < u < 18.9. May be wrong, so feel free to discuss! 
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 27072016 19:26
(Original post by AMarques)
My final answers to 1dp were 15.3 < u < 18.9. May be wrong, so feel free to discuss! 
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 27072016 19:27
(Original post by student1856)
Haha, that's right. Please tell me how you did it. Well done, by the way.
I cant be bothered to explain it in words so let me know which parts dont make sense to you, if any!
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 27072016 19:34
(Original post by student1856)
Haha, that's right. Please tell me how you did it. Well done, by the way.
Let s be the vertical displacement, t be the time, be the initial vertical velocity, and a for the acceleration. and by my previous edit (sorry I forgot the square root of 3, will edit again!), .Using the suvat equation, we get If we sub in our value of t at this given moment and rearrange, this turns the equation into . Making the subject, we get that . We know is a positive constant, so take the positive square root, to 1dp.
Scenario 2: At the goal.
This time . So using the suvat equation once again, we get . Subbing our value of the time, and rearranging we get . Hence . This will give our final result which is that (excluding the negative part of the inequality) to 1dp.Giving both of your answers, we can say that . Sorry for the wait, LaTeX takes a long time to write up. 
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 27072016 19:46
(Original post by RDKGames)
Hi, got your PM but cant send images on it from my phone so i came here :P
I cant be bothered to explain it in words so let me know which parts dont make sense to you, if any!
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 27072016 19:49
Last edited by RDKGames; 27072016 at 19:51. 
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 27072016 20:12
(Original post by RDKGames)
I skipped a step where I multiply everything by U^{2}, then factor it out after moving things around. The term to the most left would have U^{2} attached to it and so would the 3 on the other side, I moved them on the same side and moved the 1960 fraction onto the other side. Factor out the U^{2} and there we go. 
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 27072016 20:13
(Original post by AMarques)
Scenario 1: At the goalkeeper.
Let s be the vertical displacement, t be the time, be the initial vertical velocity, and a for the acceleration. and by my previous edit (sorry I forgot the square root of 3, will edit again!), .Using the suvat equation, we get If we sub in our value of t at this given moment and rearrange, this turns the equation into . Making the subject, we get that . We know is a positive constant, so take the positive square root, to 1dp.
Scenario 2: At the goal.
This time . So using the suvat equation once again, we get . Subbing our value of the time, and rearranging we get . Hence . This will give our final result which is that (excluding the negative part of the inequality) to 1dp.Giving both of your answers, we can say that . Sorry for the wait, LaTeX takes a long time to write up. 
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 27072016 20:16
(Original post by student1856)
Thank you very very much. I really appreciate this. Just checking, would this be considered a harder kinematics question? Or does it just come with practice? I've only recently started M2 and I assume you've already done it. Thanks again. 
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 28072016 11:09
(Original post by RDKGames)
This is more projectiles than normal kinematics, and it's slightly unusual from your typical questions; but you should still expect this in exams as one of the later questions. It's nothing harder because you just need practice with these projectiles and sooner or later you will be fluent enough in your methods to do these types of questions naturally. Good luck!
I'm really sorry to bother you again, but I just had a quick question. I asked some other people online about this question and they seemed to use a different method, which I don't quite understand.
Here's what they said
You can distribute the velocity of the ball: vx is constant and vx = vx0 (vx in which t = 0, x = 0)  g*tvy0 = v0cos(30), vx0 = v0sin(30)thereforex = tv0sin(30), y = tv0cos(30)  (g*t^{2)/2}when x = 10 then y > 3 therefore v0 > 15.36 m/s, when x = 25 then y < 3 therfore v0 < 18.91 m/sv0min = (50g/(10sin(30)cos(30)  3cos(30)^{2))1/2} = 15.36 m/s, v0max = (625g/(50sin(30)cos(30)  6cos(30)^{2))1/2} = 18.91 m/sI used g = 9.81 m/s^{2}Apparently he split the velocity into vertical and horizontal components and then used Pythagoras to calculate the velocity at 30 degrees.Do you understand his method? I can't seem to understand how or what he's doing in the equations above. 
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 28072016 11:24
(Original post by student1856)
Hi,
I'm really sorry to bother you again, but I just had a quick question. I asked some other people online about this question and they seemed to use a different method, which I don't quite understand.
Here's what they said
You can distribute the velocity of the ball: vx is constant and vx = vx0 (vx in which t = 0, x = 0)  g*tvy0 = v0cos(30), vx0 = v0sin(30)thereforex = tv0sin(30), y = tv0cos(30)  (g*t^{2)/2}when x = 10 then y > 3 therefore v0 > 15.36 m/s, when x = 25 then y < 3 therfore v0 < 18.91 m/sv0min = (50g/(10sin(30)cos(30)  3cos(30) = 15.36 m/s, v0max = (625g/(50sin(30)cos(30)  6cos(30) = 18.91 m/sI used g = 9.81 m/s^{2}Apparently he split the velocity into vertical and horizontal components and then used Pythagoras to calculate the velocity at 30 degrees.Do you understand his method? I can't seem to understand how or what he's doing in the equations above.
Personally, I wouldn't use the components of velocities at different points because the question talks about heights of the ball so I'm immediately thinking vertical displacement. Can you make that working out clear by use of latex? Or is that how the person said it? 
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 28072016 11:45
(Original post by RDKGames)
That is a very messy display, I've been staring at it for the last 10 mins and I still can't figure out what the hell is being done on the first line. I can understand the use of velocities to find the answer but without that working out being clearly laid out, I can't follow through with his method.
Personally, I wouldn't use the components of velocities at different points because the question talks about heights of the ball so I'm immediately thinking vertical displacement. Can you make that working out clear by use of latex? Or is that how the person said it?
EDIT: The guy said he used the s=ut+1/2at^2 for vertical and s=ut for horizontal. Then once he had the two components of velocity he used a triangle to calculate the hypotenuse velocity.Last edited by student1856; 28072016 at 11:49. 
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 28072016 12:46
(Original post by student1856)
Hi, I've written out what the guy wrote. I don't understand his method, but If you try the equations written on a calculator you get the exact right answer. Sorry, I think It uploaded sideways.
EDIT: The guy said he used the s=ut+1/2at^2 for vertical and s=ut for horizontal. Then once he had the two components of velocity he used a triangle to calculate the hypotenuse velocity. 
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 28072016 12:55
(Original post by RDKGames)
Yeah I don't understand his working out, that 0 throws me off completely, nothing is explained. If he is considering the velocities of vertical and horizontal, he should lay it out so to make it clear. I'd say just ignore it; otherwise ask him to put up his work that is clear without any awkward or needless notation.
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Updated: July 28, 2016
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