Step 1: Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations

and

Step 2: Add the horizontal and vertical components of each vector to determine the components $R_x$ and $R_y$ of the resultant vector, $\mathbf{\text{R}}$

and

Step 3: Use the Pythagorean theorem to determine the magnitude, $R$, of the resultant vector $\mathbf{\text{R}}$.

Step 4: Use a trigonometric identity to determine the direction, $\theta$, of $\mathbf{\text{R}}$.

1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position $\mathbf{s}$ are given by the quantities $x$ and $y$, and the components of the velocity $\mathbf{v}$ are given by $v_x=v\text{cos}\theta$ and $v_y=v\text{sin}\theta$, where $v$ is the magnitude of the velocity and $\theta$ is its direction.
2. Analyze the motion of the projectile in the horizontal direction using the following equations
   $$\text{Horizontal motion}(a_x=0)$$
   $$x=x_0+v_{x}t$$
   $$v_x=v_{0x}={\mathbf{\text{v}}}_{\text{x}}=\text{velocity is a constant.}$$
3. Analyze the motion of the projectile in the vertical direction using the following equations
   $$\text{Vertical motion}(\text{Assuming positive direction is up;}{a}_y=-g=-9\text{.}\text{80 m}{\text{/s}}^2)$$
   $$y=y_0+\frac{1}{2}(v_{0y}+v_y)t$$
   $$v_y=v_{0y}-\text{gt}$$
   $$y=y_0+v_{0y}t-\frac{1}{2}{\text{gt}}^2$$
   $$v_y^2=v_{0y}^2-2g(y-y_0)\text{.}$$
4. Recombine the horizontal and vertical components of location and/or velocity using the following equations
   $$s=\sqrt{x^2+y^2}$$
   $$\theta ={\text{tan}}^{-1}(y/x)$$
   $$v=\sqrt{v_x^2+v_y^2}$$
   $${\theta }_{\text{v}}={\text{tan}}^{-1}(v_y/v_x)\text{.}$$