j = MsinΘ = 25sin35 = 14.3j

**55.7m in the Y direction.**.

**-20.5i+55.7j.**

**Θ=ARCTAN(MsinΘ)/(McosΘ).**

**Note: ARCTAN means "the angle whose tangent is" and is often written as tan**

^{-1}In a soccer match, the goal keeper stands on the midpoint of her goal line. she kicks the ball 25m at an angle of 35deg to the goal line. her teammate takes the pass and kicks it 40m farther, parallel to the sideline.

a) What is the resultant displacement of the ball?

b)If the field is 110m long, how far must the next player kick the ball to take a good shot at the center of the goal, and in approximately which direction?

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This problem asks you to add 2 vectors.

One way to do this is to separate the vectors into two parts. One part is called the i-vector which is along the x axis, and the other part is the j-vector, which is along the y axis. Once the vectors are in i and j notation, you just add the i's and add the j's! Simple !

For example, if one vector is 2i + 3j, and the next one is 3i + 2j, then the sum (resultant) is (2+3)i +(3 + 2)j equals 5i + 5j.

If the vector is given to you in terms of a magnitude (size) and direction (angle) then you tear apart this vector (we say "resolve" it) into its two pieces (called components.)

There is a FORMULA for turning a vector into i and j components. If you say the magnitude is called M and the direction is called Θ, then you can write the vector as M〈Θ. (That "angle" sign is the best I can do in this editor.) You say "M at an angle of theta" when you see it written that way.

( An aside: the 'i' direction is taken to be 0 degrees and the 'j' direction is taken to be 90 degrees. This is an arbitrary choice, but for the sake of uniformity it is best to follow one convention. This is the one I choose.)

The FORMULA for turning M〈Θ into i and j values is in two parts:

i = McosΘ

j = MsinΘ

In the problem we are given, the first vector magnitude is 25, and direction is 35 degrees. 25〈35.

i = McosΘ = 25cos35= 20.5i

j = MsinΘ = 25sin35 = 14.3j

j = MsinΘ = 25sin35 = 14.3j

This vector can thus be written as 20.5 i + 14.3 j.

This means the ball traveled in the x direction (i) 20.5 units, and the y direction (j) 14.3 units.

The second vector is easier. We can see that the ball is traveling ONLY in the j direction so the vector is 40j.

This vector can thus be written as +40j.

When we add up vector (1) and vector (2) we get 20.5i + (14.3+40)j = 20.5i +54.3j.

So, we have a right triangle where we have 20.5 in the X direction and 54.3 in the y direction. Use the Pythagorean theorem to get √(20.5^2 +54.3^2) and the resultant vector MAGNITUDE is 58.04.

For part (b) of the question, first recognize that to get to the other goal line you have to travel the field length minus the distance already traveled = (110-54.3) =
**55.7m in the Y direction.**.

Since the first goalie was in the center of her goal, the second kicker must kick it the SAME number of meters back in the X direction or 20.5m. Since we called 20.5 POSITIVE the first time, we will call it NEGATIVE or (-) this time.

So the resultant vector is **-20.5i+55.7j.**

There is another FORMULA we use to solve the angle of vectors when the x and y coordinates are known (or i and j).

This FORMULA is **Θ=ARCTAN(MsinΘ)/(McosΘ).**

In this case the answer is arctan(55.7/(-20.5)) = -70 degrees. But, since the j direction is already 90 degrees, we subtract 70 from 90 and get 20 degrees.

This informs us that the 2nd players kicks the ball 20 degrees to his LEFT to hit the goal.

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