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 Ums-1. 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<======15m=====> 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'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 Ums-1. 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<======15m=====> 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.
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 usin30, we also know the acceleration due to gravity. Using the suvat equation s=ut+21at2 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 2u3t=10 hence t=u320 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!
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 usin30, we also know the acceleration due to gravity. Using the suvat equation s=ut+21at2 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 2ut=10 hence t=u20 Now the goal is 25m away, so swap that instead of 10, giving you the time it takes to reach there.Hope this helps!
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.
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.
My final answers to 1dp were 15.3 < u < 18.9. May be wrong, so feel free to discuss!
Haha, that's right. Please tell me how you did it. Well done, by the way.
Scenario 1: At the goalkeeper.
Let s be the vertical displacement, t be the time, uy be the initial vertical velocity, and a for the acceleration.s>3uy=2ua=−g and by my previous edit (sorry I forgot the square root of 3, will edit again!), t=u320.Using the suvat equation, we get 2ut−2gt2>3 If we sub in our value of t at this given moment and rearrange, this turns the equation into 3u2200g<3103−9. Making u2 the subject, we get that u2>103−9200g. We know u is a positive constant, so take the positive square root, u>15.3 to 1dp.
Scenario 2: At the goal.
This time t=u350. So using the suvat equation once again, we get 2gt2−2ut+3>0. Subbing our value of the time, and rearranging we get 3253−9<3u21250g. Hence u2<253−91250g. This will give our final result which is that (excluding the negative part of the inequality) u<18.9 to 1dp.Giving both of your answers, we can say that 15.3<u<18.9. Sorry for the wait, LaTeX takes a long time to write up.
I skipped a step where I multiply everything by U2, then factor it out after moving things around. The term to the most left would have U2 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 U2 and there we go.
I skipped a step where I multiply everything by U2, then factor it out after moving things around. The term to the most left would have U2 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 U2 and there we go.
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.
Let s be the vertical displacement, t be the time, uy be the initial vertical velocity, and a for the acceleration.s>3uy=2ua=−g and by my previous edit (sorry I forgot the square root of 3, will edit again!), t=u320.Using the suvat equation, we get 2ut−2gt2>3 If we sub in our value of t at this given moment and rearrange, this turns the equation into 3u2200g<3103−9. Making u2 the subject, we get that u2>103−9200g. We know u is a positive constant, so take the positive square root, u>15.3 to 1dp.
Scenario 2: At the goal.
This time t=u350. So using the suvat equation once again, we get 2gt2−2ut+3>0. Subbing our value of the time, and rearranging we get 3253−9<3u21250g. Hence u2<253−91250g. This will give our final result which is that (excluding the negative part of the inequality) u<18.9 to 1dp.Giving both of your answers, we can say that 15.3<u<18.9. Sorry for the wait, LaTeX takes a long time to write up.
I really appreciate your help, thank you very much. Have a great day.
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.
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!
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!
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*t2)/2when 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/s2Apparently 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.
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*t2)/2when 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/s2Apparently 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.
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?
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?
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.
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.
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.
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.
I've asked him to elaborate on his working. I'll get back to you when he replies.