Answer
Horizontal component $=20\sqrt{3}$
Vertical component $=20$
$\vec{v}=20\sqrt{3}\vec{i}+20\vec{j}$
Work Step by Step
$|\vec{v}|=40$ $,$ $\theta=30^{\circ}$
Multiply the magnitude of the vector $\vec{v}$ by the cosine of $\theta$ to obtain its horizontal component:
Horizontal component $=|\vec{v}|\cos\theta=40\cos30^{\circ}=40\Big(\dfrac{\sqrt{3}}{2}\Big)=20\sqrt{3}$
Multiply the magnitude of the vector $\vec{v}$ by the sine of $\theta$ to obtain its vertical component:
Vertical component $=|\vec{v}|\sin\theta=40\sin30^{\circ}=40\Big(\dfrac{1}{2}\Big)=20$
Write the vector in terms of $\vec{i}$ and $\vec{j}$. Do so by adding the product between its horizontal component and $\vec{i}$ and the product between its vertical component and $\vec{j}$:
$\vec{v}=20\sqrt{3}\vec{i}+20\vec{j}$
