Ahiker's trip consists of three segments. path a is 9.0 km long heading 60.0° north of east. path b is 6.0 km long in a direction due east. path c is 3.0 km long heading 315° counterclockwise from east.

(a) graphically add the hiker's displacements in the order a, b, c.
magnitude of km
direction of ° (counterclockwise from east)

(b) graphically add the hiker's displacements in the order c, b, a.
magnitude of
direction of ° (counterclockwise from east)

(c) what can you conclude about the resulting displacements?

magnitud = 10.9 km    θ = 15º.   the displacements are equal

Explanation:

The sum of vectors can be done by two methods: one graphic and one analytical

In the graphic method, the first vector is drawn, then the second vector is drawn on the end of it and so on, the vector is finely drawn resulting from the origin of the first to the tip of the last. Look at the attachments for our case

When you examine the drawings, you conclude that the displacements are equal, this is because of the commutative property of the addition

a) magnitud = 13.8 km    θ = 24º

b) magnitud =13.8 km    θ = 24º

analytical method

To find the values, we decompose the vectors, using trigonometry

Ax = 9 cos 60 = 4.5 km

Ay = 9 sin 60  = 7.8 km

Bx = 6 km

Cx = 3 cos 315 = 2.12 km

Cy = 3 sin 315 = -2.12 km

We calculate the result on each axis

Rx = Ax + Bx + Cx

Ry = Ay + By + Cy

Rx = 4.5 + 6 + 2.12

Rx = 12.62 km

Ry = 7.8 +0 -2.12

Ry = 5.68 km

Using Pythagoras and trigonometry we build the resulting vector

R = √(12.62² +5.68²)

R = 13.84 km

tan θ = Ry / Rx

θ = tan⁻¹ (5.68 / 12.62)

T = 24º

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be