How do you solve a projectile motion problem?

Usually, a projectile is launched at an angle. Your very first step is to split that initial velocity ($v_0$) into its horizontal and vertical pieces using trigonometry. If the object is launched at an angle $\theta$ from the ground, the initial horizontal velocity is $v_{0x} = v_0 \cos \theta$ and the initial vertical velocity is $v_{0y} = v_0 \sin \theta$. Once you do this, you can put the original angled velocity away. You will only use the $x$ and $y$ components from this point forward.

To keep from mixing up your numbers, create an X-column and a Y-column. In the X-column, there is no acceleration ($a_x = 0$), so the only equation you need is $x = v_{0x} t$. In the Y-column, gravity is constantly pulling down, so the acceleration is $a_y = -9.8 \text{ m/s}^2$ (or $-g$). Here, you will use the standard kinematic equations, like $y = v_{0y}t - \frac{1}{2}gt^2$ and $v_y = v_{0y} - gt$. Never put an X-value into a Y-equation, or vice versa!

The horizontal and vertical motions are independent, but they share one crucial variable: time ($t$). The object hits the ground at the exact same moment in both the X and Y directions. Usually, you will use the Y-column to figure out how long the object is in the air (because gravity dictates when it hits the ground), and then plug that time into the X-column to see how far it traveled horizontally.

The most common mistake is assuming that the velocity is zero at the very top of the projectile's arc. While it is true that the object stops moving upward for a split second (meaning $v_y = 0$), it is still moving forward! The horizontal velocity ($v_x$) at the highest point is exactly the same as it was when it was launched. Always remember that the overall velocity is only completely zero if the object has hit a wall or the ground.