Notices
Results 1 to 27 of 27

Thread: shooting on a motorway

  1. #1 shooting on a motorway 
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    2 cars, A,B start from O at T_0 on two roads, A points North (0°) B at 60° (300°) at speed v
    After 10 s (at distance 10v), a shot is fired trying to hit B. The bullet has speed v.

    If there is a solution,, at what angle must the shot be fired?


    Reply With Quote  
     

  2.  
     

  3. #2  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by whizkid View Post
    2 cars, A,B start from O at T_0 on two roads, A points North (0°) B at 60° (300°) at speed v
    After 10 s (at distance 10v), a shot is fired trying to hit B. The bullet has speed v.

    If there is a solution,, at what angle must the shot be fired?
    yes


    Reply With Quote  
     

  4. #3  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    Quote Originally Posted by xyzt View Post
    yes
    Are you sure? have you really found a solution?
    Reply With Quote  
     

  5. #4  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by xyzt View Post
    yes
    Are you sure? have you really found a solution?
    Write the equations of motion and you will find out. Talk is cheap.
    Reply With Quote  
     

  6. #5  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    According to my calc, there is no solution because the inertia of the bullet points at 0° and if you point the gun at the car B (240°, 120 ° to the left of road A) the trajectory of the bullet is parallel to road B and speed exactly = v, or ahead of the car (<120° to the left od A) it will diverge from road B toward road A. If you point it behind (>120°) the resulting vector crosses B but only after the car has passed, since speed is >v.
    In no case B will be hit.

    If you can't be bothered to discuss it,give me just the value of your solution: angle and time, and I'll do reverse engineering

    Thanks for you time
    Reply With Quote  
     

  7. #6  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by whizkid View Post
    According to my calc,
    Let's see your "calcs".


    there is no solution because the inertia of the bullet points at 0°
    This exercise has nothing to do with "inertia', it is a simple problem of kinematics.



    If you can't be bothered to discuss give me just the value of your solution angle and time.

    Thanks for you time
    We don't do your homework in this forum. Show your calcs" and we'll talk.
    Reply With Quote  
     

  8. #7  
    Suspended
    Join Date
    Apr 2007
    Location
    Pennsylvania
    Posts
    8,795
    The cars are going at the same speed as the bullet?
    Reply With Quote  
     

  9. #8  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    Quote Originally Posted by Harold14370 View Post
    The cars are going at the same speed as the bullet?
    Yes, v_cars=v_bullet =v, let's put v =1m/s for the sake of semplicity: distance A,B is 10 m, and the bullet, after the shot, keeps moving at 1m/s North (direction: 0°). If we aim at car B, we add a vector at 120° from the other vector (direction: 240°).
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by whizkid View Post
    According to my calc,
    This exercise has nothing to do with "inertia', it is a simple problem of kinematics.
    We don't do your homework in this forum. Show your calcs" and we'll talk.
    This is not homework, and if you do not realize that on the bullet act 2 kinetic vectors it is useless to discuss it with you.
    I fully and clearly argumented my analysis, try to understand it and if there is a mistake, point it out.
    Reply With Quote  
     

  10. #9  
    Forum Professor jrmonroe's Avatar
    Join Date
    Mar 2010
    Posts
    1,444
    Rewrite the problem so Car A is stationary and Car B has acquired the opposite velocity as Car A originally had in addition to Car B's own velocity. Then the nature of the situation will become clearer. Of course, this will cancel various aerodynamic effects on the bullet, but it's probably okay to overlook them, considering the simple nature of the problem.
    Grief is the price we pay for love. (CM Parkes) Our postillion has been struck by lightning. (Unknown) War is always the choice of the chosen who will not have to fight. (Bono) The years tell much what the days never knew. (RW Emerson) Reality is not always probable, or likely. (JL Borges)
    Reply With Quote  
     

  11. #10  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    Quote Originally Posted by jrmonroe View Post
    Rewrite the problem so Car A is stationary and Car B has acquired the opposite velocity as Car A originally had in addition to Car B's own velocity. Then the nature of the situation will become clearer. Of course, this will cancel various aerodynamic effects on the bullet, but it's probably okay to overlook them, considering the simple nature of the problem.
    Of course we are ignoring air friction.
    Yo can't do that, because you would cancel the fact that the bullet keeps moving in the direction of road A at 1 m/s, after it has been fired.
    Reply With Quote  
     

  12. #11  
    Suspended
    Join Date
    Apr 2007
    Location
    Pennsylvania
    Posts
    8,795
    At the time the shot is fired, A, B and O form an isosceles triangle with AO and BO being of length = 10v. Since angle AOB is 60 degrees, the other two angles are also 60 degrees, and it is an equilateral triangle. If the bullet intersects the path of car B at point I, there is a triangle AIB. Since the bullet has the same speed as B, then AI would have to be equal to BI, but that can't happen because angle ABI is 120 degrees.
    Reply With Quote  
     

  13. #12  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by whizkid View Post
    According to my calc,
    This exercise has nothing to do with "inertia', it is a simple problem of kinematics.
    We don't do your homework in this forum. Show your calcs" and we'll talk.
    This is not homework, and if you do not realize that on the bullet act 2 kinetic vectors it is useless to discuss it with you.
    This is meaningless gibberish.


    I fully and clearly argumented my analysis, try to understand it and if there is a mistake, point it out.
    I see, you haven't done any calculations, so "according to your calcs" is nothing but a pretense.
    Reply With Quote  
     

  14. #13  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by Harold14370 View Post
    Since the bullet has the same speed as B, then AI would have to be equal to BI, but that can't happen because angle ABI is 120 degrees.
    He means that A,B have speed v wrt the ground. The bullet has speed v wrt car A. Then , the scenario is possible. Frankly, I think that he doesn't even understand the scenario.
    Reply With Quote  
     

  15. #14  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    The only one who doesn't understand here, it's you. An intelligent reader needs no calcs /equations, just some brain

    There is no solution both if v is the power of fire of the gun (at rest) and if v is the effective velocity of the bullet
    Reply With Quote  
     

  16. #15  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by whizkid View Post
    The only one who doesn't understand here, it's you. An intelligent reader needs no calcs /equations, just some brain

    There is no solution both if v is the power of fire of the gun (at rest) and if v is the effective velocity of the bullet
    This is just more gibberish. Why don't you scan the problem statement and post it? At least, this way, we can have a coherent problem statement. Yours isn't coherent.
    Reply With Quote  
     

  17. #16  
    Moderator Moderator Janus's Avatar
    Join Date
    Jun 2007
    Posts
    2,223
    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by jrmonroe View Post
    Rewrite the problem so Car A is stationary and Car B has acquired the opposite velocity as Car A originally had in addition to Car B's own velocity. Then the nature of the situation will become clearer. Of course, this will cancel various aerodynamic effects on the bullet, but it's probably okay to overlook them, considering the simple nature of the problem.
    Yo can't do that, because you would cancel the fact that the bullet keeps moving in the direction of road A at 1 m/s, after it has been fired.
    Yes, you can. If v, the velocity of the bullet is with respect to the car, then all you have to do is work out the resultant vector of car B with respect to A. Since A and B started at the same point and are both traveling at the same speed with respect to that point, then the relative direction of B with respect to A never changes (it will always be 120 degrees from the direction A faces). So A just has to aim directly at B. However, Since as already pointed out, the origin, A and B form a equilateral triangle, B will be at a distance of 10v and receding at v with respect to A. Since the bullet travels at v wrt to A, it can never catch B. The bullet's velocity with respect to the road is irrelevant.

    If however, if the velocity of the bullet is v wrt to the road, then it is easier to use the road as the frame of reference. In this case, the actual velocity of A at the time the bullet is fired is irrelevant. The situation would be exactly the as if, instead of firing the bullet from A, the bullet is fired by some on the road the instant A passes him.( granted, the actual muzzle velocity and direction fired for the gun in the car would be different, but we can cross that bridge if we need to later.)

    What we are trying to solve for is a triangle with the three points being: Where the bullet(A) is fired from, B's position at this time(B), and the point where the the bullet and B meet(C).
    We know that side AB is 10v, We also know that angle ABC is 120 degree. And since the speed of both the bullet and B is v, both sides BC and AC must be equal (=xv) It should be obvious at this point that such a triangle can't exist. (angles ABC and BAC would have to be equal and together they would sum to greater than 180 degrees)
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


    Edit/Delete Message
    Reply With Quote  
     

  18. #17  
    Moderator Moderator
    Join Date
    Apr 2007
    Location
    Washington State
    Posts
    8,416
    Quote Originally Posted by whizkid View Post
    There is no solution both if v is the power of fire of the gun (at rest) and if v is the effective velocity of the bullet
    You don't seem to know what the meaning of "power" or "velocity" means. Since this is the hard science part of the forum, you'd should be a lot more careful about their use or expect a thread move to general.

    Have you done a sketch with the appropriate angles and movement vectors? A lot of the points being made, and the solution less confusing if you did.
    Meteorologist/Naturalist & Retired Soldier
    “The Holy Land is everywhere” Black Elk
    Reply With Quote  
     

  19. #18  
    Forum Professor jrmonroe's Avatar
    Join Date
    Mar 2010
    Posts
    1,444
    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by jrmonroe View Post
    Rewrite the problem so Car A is stationary and Car B has acquired the opposite velocity as Car A originally had in addition to Car B's own velocity. Then the nature of the situation will become clearer. Of course, this will cancel various aerodynamic effects on the bullet, but it's probably okay to overlook them, considering the simple nature of the problem.
    Yo can't do that, because you would cancel the fact that the bullet keeps moving in the direction of road A at 1 m/s, after it has been fired.
    Oops. Ok, also add the opposite velocity of Car A to the bullet. Basically add the opposite velocity of Car A to the entire system, which makes Car A stationary and changes the velocity of all other objects. Then solve.
    Grief is the price we pay for love. (CM Parkes) Our postillion has been struck by lightning. (Unknown) War is always the choice of the chosen who will not have to fight. (Bono) The years tell much what the days never knew. (RW Emerson) Reality is not always probable, or likely. (JL Borges)
    Reply With Quote  
     

  20. #19  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    If someone has found a solution, it would be wiser to present it, instead of arguing.
    I do not see the reason why someone wants to complicate matters.
    The problem is very simple :

    at the moment of the shot the distance between the cars D is 10m, and it is growing at the rate of 1m /s.
    In order to catch up with the distance, the speed of the bullet must be > than 1m/s if it is =1 it will never hit car B.

    If the speed of the bullet is the actual speed (v_a=1) or is the speed when the gun is at rest (v_b=1), doesn't change the result, because in the second case v_a (the sum of the speed of car A =1 and the bullet =1) varies 0<v_a<2 , but when the actual speed v_a >1 the direction is diverging from road B.

    If this is not right , correct the mistake, please.
    Last edited by whizkid; June 22nd, 2014 at 01:32 AM.
    Reply With Quote  
     

  21. #20  
    Forum Freshman
    Join Date
    Mar 2014
    Posts
    65
    Quote Originally Posted by whizkid View Post
    According to my calc, there is no solution because the inertia of the bullet points at 0° and if you point the gun at the car B (240°, 120 ° to the left of road A) the trajectory of the bullet is parallel to road B and speed exactly = v, or ahead of the car (<120° to the left od A) it will diverge from road B toward road A. If you point it behind (>120°) the resulting vector crosses B but only after the car has passed, since speed is >v.
    In no case B will be hit.
    Let me get this straight: if fired at 120º the bullet would travel at 60º with a net velocity over car A of +41.5% and -3.4% for a total velocity of 1.38v. This would pass up the car B at 1v before T=40.

    Why do you think that the vector sum is just v?

    Increasing the angle fired would cause a path intercept. Hitting car B is a matter of converging B distance+10 with bullet distance @ a vector velocity in the same timeframe.
    Reply With Quote  
     

  22. #21  
    Forum Bachelors Degree
    Join Date
    Apr 2011
    Location
    U.S.A
    Posts
    414
    We're not going to help you kill somebody.
    With bravery and recognition that we are harbingers of our destiny and with a paragon of virtue.
    Reply With Quote  
     

  23. #22  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    Quote Originally Posted by G O R T View Post

    Let me get this straight: if fired at 120º the bullet would travel at 60º with a net velocity over car A of +41.5% and -3.4% for a total velocity of 1.38v. This would pass up the car B at 1v before T=40.
    Why do you think that the vector sum is just v?
    .
    If you aim at 120° you are pointing straight at car B, if you add 2 vectors =1 at 120° the vector sum is =1 and points at 60°, the sum 1.41 occurs when the angle is 90°, right?
    Reply With Quote  
     

  24. #23  
    Forum Bachelors Degree One beer's Avatar
    Join Date
    May 2013
    Posts
    442
    2 cars, A,B start from O at T_0 on two roads, A points North (0°) B at 60° (300°) at speed v
    After 10 s (at distance 10v), a shot is fired trying to hit B. The bullet has speed v.


    If there is a solution,, at what angle must the shot be fired?
    We need a diagram, because although you might understand what is in your head, you have not explained the question very clearly to us.

    Car A travels North.

    Car B travels which way? - you say "60° (300°)", what does that mean?

    Where is the bullet shot from?


    OB

    PS, don't have a go at people who are trying to help you.
    Reply With Quote  
     

  25. #24  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by whizkid View Post

    If the speed of the bullet is the actual speed (v_a=1) or is the speed when the gun is at rest (v_b=1), doesn't change the result, because in the second case v_a (the sum of the speed of car A =1 and the bullet =1) varies 0<v_a<2 , but when the actual speed v_a >1 the direction is diverging from road B.
    This is, of course, false. It is important to specify if the speed of the bullet is referenced to the ground or to the moving car A. Speeds do not add up, velocities do. Based on your error above, you seem not to know this fact.
    I asked you to scan the original problem statement and to post it here, why aren't you doing that?
    Reply With Quote  
     

  26. #25  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by One beer View Post
    2 cars, A,B start from O at T_0 on two roads, A points North (0°) B at 60° (300°) at speed v
    After 10 s (at distance 10v), a shot is fired trying to hit B. The bullet has speed v.


    If there is a solution,, at what angle must the shot be fired?
    We need a diagram, because although you might understand what is in your head, you have not explained the question very clearly to us.

    Car A travels North.

    Car B travels which way? - you say "60° (300°)", what does that mean?

    Where is the bullet shot from?


    OB

    PS, don't have a go at people who are trying to help you.
    Car A travels along the x axis, The gun is located in car A
    Car B travels along a line inclined 60 degrees wrt the positive x axis, in the same direction as car A.
    At any moment in time, the triangle OAB is equilateral, as pointed out by Janus.
    The line AB forms an angle of 120 degrees with the positive x axis. The gun needs to be inclined at an angle LESS than 120 degrees wrt to the x axis, such that the bullet intercepts B. This is a simple problem of artillery.
    Last edited by Howard Roark; June 22nd, 2014 at 09:15 AM.
    Reply With Quote  
     

  27. #26  
    Suspended
    Join Date
    Apr 2007
    Location
    Pennsylvania
    Posts
    8,795
    Janus explained very well why the bullet cannot hit the target, whether v is ground speed or muzzle velocity. Why is there still a debate about it?
    Reply With Quote  
     

  28. #27  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by Harold14370 View Post
    Janus explained very well why the bullet cannot hit the target, whether v is ground speed or muzzle velocity. Why is there still a debate about it?
    I think the part where the speed of the bullet is given wrt the car is incorrect in Janus' analysis. My analysis shows that contrary to his conclusion, the bullet can hit B. I haven't posted the math yet because I want to see the actual problem statement from whizzkid, something that he has not done yet.
    I agree with Janus' analysis for the case where all speeds are given wrt ground.
    Reply With Quote  
     

Similar Threads

  1. Speed of Light - Shooting Star
    By teynon in forum Astronomy & Cosmology
    Replies: 8
    Last Post: November 29th, 2012, 01:40 PM
  2. The Colorado Shooting
    By tszy in forum Behavior and Psychology
    Replies: 12
    Last Post: July 27th, 2012, 12:24 PM
  3. The Science of Shooting String
    By CuriousFool in forum Mechanical, Structural and Chemical Engineering
    Replies: 3
    Last Post: August 15th, 2010, 01:06 PM
  4. US gets its fourth school shooting in the last week
    By Demen Tolden in forum Criminology and Forensic Science
    Replies: 14
    Last Post: March 29th, 2009, 08:36 PM
  5. anyone know how fast is a shooting bullet?thnx
    By lovenara in forum Military Technology
    Replies: 7
    Last Post: October 7th, 2007, 06:00 PM
Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •