Question

Vector A has y-component Ay= +15.0 m . A makes an angle of 32.0 counterclockwise from the +y-axis. What is the x component of A? What is the magnitude of A?

Answer 1

Answer:

x-component=-9.3 m

Magnitude of A=17.7m

Explanation:

We are given that

$A_y=+15 m$

$\theta=32^{\circ}$

We have to find the x-component of A and magnitude of A.

According to question

$A_y=\mid A\mid cos\theta$

Substitute the values then we get

$15=\mid A\mid cos32$

$\mid A\mid =\frac{15}{cos32}=\frac{15}{0.848}$

$\mid A\mid=17.7$m

$tan\theta=\frac{perpendicular\;side}{Base}$

$tan32=\frac{A_x}{A_y}=\frac{A_x}{15}$

$0.62\times 15=A_x$

$A_x=9.3$

The value of x-component of A is negative because the vector A lie  in second quadrant.

Hence, the x- component of A=-9.3 m

Answer 2

Answer:

The x component is 24.0 m and the magnitude of vector A is 28.30 m.

Explanation:

Given that,

Vertical component $A_{y}=+15.0 m$

Angle = 32.0

We need to calculate the x component of A

Using formula of angle

$\tan\theta=\dfrac{A_{y}}{A_{x}}$

$A_{x}=\dfrac{A_{y}}{\tan\theta}$

Where, $A_{y}$ = vertical component

$A_{x}$ =  horizontal component

Put the value into the formula

$A_{x}=\dfrac{15.0}{\tan32}$

$A_{x}=24.0\ m$

We need to calculate the magnitude of A

Using formula of vector magnitude

$|A|=\sqrt{(15.0)^2+(24.0)^2}$

$|A|=28.30\ m$

Hence, The x component is 24.0 m and the magnitude of vector A is 28.30 m.