Question

Find the vertical and horizontal component ? Steps please

Answer 1

`V_H = 8.62` and `V_V = 11.03` to 2 dp

Explanation:

This is just trigonometry. We know that

`sin(theta) = "opposite"/"hypotenuse"` and `cos(theta) = "adjacent"/"hypotenuse"`

So for all problems like this we can instantly see that the opposite corresponds to the vertical and the adjacent corresponds to the horizontal.

`V_H = abs(vec V)cos(theta)` and `V_V = abs(vec V) sin(theta)`

In this case, the magnitude is 14 and `theta = 52.0`

`V_H = 14*cos(52) = 8.62` and `V_V = 14*sin(52) = 11.03`

Answer 2

`V_v approx 11.032` (3 decimal places)
`V_h approx 8.619` (3 decimal places)

Explanation:

You may recall SOHCAHTOA for a right angled triangle from trigonometry:

H = Hypotenuse, O = Opposite side, A = Adjacent side

`sin(theta) = (O)/(H), cos(theta) = A/H, tan(theta) = (O)/(A)`

In this case, the horizontal component `V_h` is adjacent to the angle, and the vertical component `V_v` is opposite the angle. In both cases, `theta` is given as `52°` and the hypotenuse is given as `14.0`.

We can note the two relationships to solve the vector into components:

`sin(52) = (O)/H = (V_v)/14`

`cos(52) = A/H = (V_h)/14`

rearranging for `V_v` and `V_h`,

`V_v = sin(52)*14 approx 11.03215055 approx 11.032` (3 decimal places)
`V_h = cos(52)*14 approx 8.619260655 approx 8.619` (3 decimal places)