vector's horizontal component?

Question

Given vector Y = 73.1 m ∠ − 40.0° , find its horizontal component

Answer 1

73.1 cos 40° = 56 m (to nearest whole number)

Answer 2

this means that the vector is 73.1 m. to find its components, we will use trigonometry. why? simple because we will need to find a relationship between your given length (which in this case is 73.1) and the component you are asked (particularly the horizontal component).


the horizontal component is simply given by the quantity of the vector multiplied to the cosine function of the angle. In symbols:

Horizontal component = y*cos (angle)


The same goes for the Vertical component:

Vertical component = y* sin (angle)

Thus:
H = y*cos$
H=73.1*cos(-40)
H= 56.998 m


If you are interested in proving the formula I gave you, read further.

imagine a right triangle with hypotenuse of length 73.1. graphically, your vector is the hypotenuse. the opposite side (relative to the 40 deg) is the vertical component. and the adjacent side is the horizontal component.

recall:
sin $ = opposite/hypotenuse = vertical component / vector
cos $ = adjacent/hypotenuse= horizontal component/vector

cross multiplying each equation, you get:
vertical component = vector * sin$
horizontal component = vector * cos$

Answer 3

This means that the magnitude of the vector is 73.1m and the vector is making an angle of -40 degrees with the positive direction of x-axis.This means that the vector is 40 degrees below the positive direction of x-axis.
Horizontal component=73.1Xcos (angle which the vector is making with the x-axis)
=73.1Xcos (40 degrees)
=73.1X0.77
=56.29m(approx).
Be assured of the answer.I am an engineering student and I practice all these daily.

Answer 4

No. V^2=h^2+v^2 so h and v must be smaller than V

Answer 5

y=73m
angle =-40
horizontal component=y*sin 40

Answer 6

Form a triangle, and apply Pitagora.

Answer 7

horiz. comp. =Ycos40 = 73.1cos40 = 55.997 = 56m approx.