Question

what is the horizontal component of a vector that has a magnitude of 5 and is at an angle of 38 degrees?
3.08
33.5
33
3.94
43
what is the vertical component of a vector that has a magnitude of 5 and is at an angle of 38 degrees?
33.5
3.94
3.08
33
43

Explanation:

Step1: Recall vector - component formula

The horizontal component of a vector $\vec{v}$ with magnitude $|\vec{v}|$ and angle $\theta$ with the horizontal is given by $v_x = |\vec{v}|\cos\theta$.

Step2: Substitute values

Given $|\vec{v}| = 5$ and $\theta=38^{\circ}$, we have $v_x = 5\times\cos(38^{\circ})$.
Since $\cos(38^{\circ})\approx0.788$, then $v_x = 5\times0.788 = 3.94$.
The vertical component of a vector $\vec{v}$ with magnitude $|\vec{v}|$ and angle $\theta$ with the horizontal is given by $v_y=|\vec{v}|\sin\theta$.
Substitute $|\vec{v}| = 5$ and $\theta = 38^{\circ}$, since $\sin(38^{\circ})\approx0.616$, then $v_y=5\times0.616 = 3.08$.

Answer:

1. For the horizontal - component question: D. 3.94
2. For the vertical - component question: A. 3.08