# Thread: shooting on a motorway

1. 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?

2.

3. Originally Posted by whizkid
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

4. Originally Posted by xyzt
yes
Are you sure? have you really found a solution?

5. Originally Posted by whizkid
Originally Posted by xyzt
yes
Are you sure? have you really found a solution?
Write the equations of motion and you will find out. Talk is cheap.

6. 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

7. Originally Posted by whizkid
According to my calc,

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.

8. The cars are going at the same speed as the bullet?

9. Originally Posted by Harold14370
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°).
Originally Posted by xyzt
Originally Posted by whizkid
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.

10. 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.

11. Originally Posted by jrmonroe
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.

12. 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.

13. Originally Posted by whizkid
Originally Posted by xyzt
Originally Posted by whizkid
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.

14. Originally Posted by Harold14370
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.

15. 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

16. Originally Posted by whizkid
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.

17. Originally Posted by whizkid
Originally Posted by jrmonroe
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)

18. Originally Posted by whizkid
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.

19. Originally Posted by whizkid
Originally Posted by jrmonroe
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.

20. 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.

21. Originally Posted by whizkid
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.

23. Originally Posted by G O R T

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?

24. 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.

25. Originally Posted by whizkid

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?

26. Originally Posted by One beer
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.

27. 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?

28. Originally Posted by Harold14370
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.

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