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Intersection of two Objects

  • Thread starter cloj63
  • Start date
  • #1
2
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Homework Statement



Ok, I need to find the angle at which I will throw an object (O1) so that it intersect with another one (O2) already in motion.

I know the speed of each object (constant speed, no acceleration), I know the angle of O2 and its distance from me at time=0

Homework Equations



Speed O1 10u/s
Speed O2 3u/s

Angle of O1: unknown
Angle of O2: ∏/3

Position of O2 at t0: 40 units, at an angle of 3∏/2 in O1 perspective

The Attempt at a Solution



THIS ATTEMPT WENT TO NOTHING, IT IS LONG AND MAYBE CONFUSING SAID LIKE THAT IN A FORUM... read if you prefer. In reality; I am searching for a way to solve the problem that is not this one (as I get negative discriminant in a quadratic thus impossible to solve). What path would you use to solve it?

Here is my first attempt to solve it; google doc sheets and its written in french however:
https://docs.google.com/file/d/0BzKB0-4y9gXPR0dzVGc2elVUUC1ud2Z4VWdjdkpOdw/edit?pli=1

There is a lot of errors in that, and the first half page is just junk and irrelevant in the end.

To summarize: I think it as a triangle problem where P1 is me, P2 is the other object and P3 is the intersection point.
I know one angle: P2 because I know O2 position and its angle of motion [∏/2-(2∏-3∏/2)]+[∏/2-∏/3]=∏/6.
I also know the P1P2 side; 40 units
I don't exactly know the other two sides, but I know their relation to each other:

P1P3 side is the distance my object will move: P1P3=t*(Speed of O1)=t*10u/s

P2P3 side is the distance the other object will move: P2P3=t*(Speed of O2)=t*3u/s

P3 is the point they will intersect, thus at P3; t of both will be equal. t=P2P3/(3u/s)
Thus I can substitute in one equation: P1P3=P2P3/(3u/s)*10u/s

Long story short; I was able to isolate what my Angle P1 was, but it required using Law of Cosin and a quadratic equality... to get a very long equation (you can see the derivation starting from middle 2nd page if you want). But the quadratic left a radical, and unfortunately most scenario I am in lead to a negative answer IN the quadratic (Discriminant <0) and I can't understand why it could give me an impossible scenario in an intersection problem...
 

Answers and Replies

  • #2
10
0

Homework Statement



Ok, I need to find the angle at which I will throw an object (O1) so that it intersect with another one (O2) already in motion.

I know the speed of each object (constant speed, no acceleration), I know the angle of O2 and its distance from me at time=0

Homework Equations



Speed O1 10u/s
Speed O2 3u/s

Angle of O1: unknown
Angle of O2: ∏/3

Position of O2 at t0: 40 units, at an angle of 3∏/2 in O1 perspective

The Attempt at a Solution



THIS ATTEMPT WENT TO NOTHING, IT IS LONG AND MAYBE CONFUSING SAID LIKE THAT IN A FORUM... read if you prefer. In reality; I am searching for a way to solve the problem that is not this one (as I get negative discriminant in a quadratic thus impossible to solve). What path would you use to solve it?

Here is my first attempt to solve it; google doc sheets and its written in french however:
https://docs.google.com/file/d/0BzKB0-4y9gXPR0dzVGc2elVUUC1ud2Z4VWdjdkpOdw/edit?pli=1

There is a lot of errors in that, and the first half page is just junk and irrelevant in the end.

To summarize: I think it as a triangle problem where P1 is me, P2 is the other object and P3 is the intersection point.
I know one angle: P2 because I know O2 position and its angle of motion [∏/2-(2∏-3∏/2)]+[∏/2-∏/3]=∏/6.
I also know the P1P2 side; 40 units
I don't exactly know the other two sides, but I know their relation to each other:

P1P3 side is the distance my object will move: P1P3=t*(Speed of O1)=t*10u/s

P2P3 side is the distance the other object will move: P2P3=t*(Speed of O2)=t*3u/s

P3 is the point they will intersect, thus at P3; t of both will be equal. t=P2P3/(3u/s)
Thus I can substitute in one equation: P1P3=P2P3/(3u/s)*10u/s

Long story short; I was able to isolate what my Angle P1 was, but it required using Law of Cosin and a quadratic equality... to get a very long equation (you can see the derivation starting from middle 2nd page if you want). But the quadratic left a radical, and unfortunately most scenario I am in lead to a negative answer IN the quadratic (Discriminant <0) and I can't understand why it could give me an impossible scenario in an intersection problem...
X = the distance (X-axis) from origin of the final position of both objects when they meet.
Y = the distance (Y-axis) from origin of the final position of both objects when they meet.
y1 = distance traveled by object 1 in the Y-axis
y2 = distance traveled by object 2 in the Y-axis
x1 = distance traveled by object 1 in the X-axis
x2 = distance traveled by object 2 in the X-axis
X1i = initial position of object 1 in the X-axis
X2i = initial position of object 2 in the X-axis
Y1i = initial position of object 1 in the Y-axis
Y2i = initial position of object 2 in the Y-axis

X = x1 + X1i = x2 + X2i
Y = y1 + Y1i = y2 + Y2i

initial position of both is already given. i.e X1i X2i Y1i Y2i

x1 = (V1 * Cos (θ1)) * t
x2 = (V2 * Cos (θ2)) * t
y1 = (V1 * Sin (θ1)) * t
y2 = (V2 * Sin (θ2)) * t



Now good luck.
 
  • #3
2
0
thanks, I will try this out
 

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