Question

1. consider a vector 5 units long at an angle 53° to the horizontal. a. determine the horizontal component of the vector. b. determine the vertical component of the vector.

Explanation:

Step1: Recall vector - component formula

The horizontal component of a vector $\vec{v}$ with magnitude $v$ and angle $\theta$ with the horizontal is given by $v_x = v\cos\theta$, and the vertical component is given by $v_y=v\sin\theta$. Here $v = 5$ and $\theta = 53^{\circ}$.

Step2: Calculate the horizontal component

We know that $\cos(53^{\circ})\approx0.6$. Using the formula $v_x = v\cos\theta$, we substitute $v = 5$ and $\theta = 53^{\circ}$. So $v_x=5\times\cos(53^{\circ})=5\times0.6 = 3$.

Step3: Calculate the vertical component

We know that $\sin(53^{\circ})\approx0.8$. Using the formula $v_y = v\sin\theta$, we substitute $v = 5$ and $\theta = 53^{\circ}$. So $v_y=5\times\sin(53^{\circ})=5\times0.8 = 4$.

Answer:

a. The horizontal component of the vector is 3.
b. The vertical component of the vector is 4.