How Do You Solve Projectile Motion Problems (Horizontal and Vertical Components)?

Resolve the initial velocity into horizontal and vertical components, then analyse each direction independently using the constant-acceleration equations. The two motions share the same time but have different accelerations.

The Method

In IB Physics A.1 Kinematics (SL and HL), ideal projectile motion assumes negligible fluid resistance. Choose positive directions first: usually right and upward, and resolve an initial velocity $u$ launched at angle $\theta$:

$$u_x=u\cos\theta, \qquad u_y=u\sin\theta$$

Apply a suitable equation of motion separately in each direction, such as

$$s=ut+\frac{1}{2}at^2, \qquad v=u+at, \qquad v^2=u^2+2as.$$

Why the Components Can Be Separated

Gravity acts vertically, so it changes only the vertical component of velocity. With no fluid resistance, there is no horizontal resultant force and therefore no horizontal acceleration. This independence does not mean two separate projectiles exist: the components combine vectorially at every instant to give the actual velocity, $v=\sqrt{v_x^2+v_y^2}$, and its direction follows from .

Connect the components using time, which is identical for both motions. Keep signs consistent throughout. With upward positive, vertical displacement, vertical velocity, and acceleration may be negative; this does not automatically mean the projectile is slowing down. It slows while rising and speeds up while falling because acceleration remains downward.

For example, a projectile is launched at $20\,\text{m s}^{-1}$ at $30^\circ$ and lands at its launch height:

$$u_x=20\cos30^\circ=17.3\,\text{m s}^{-1}, \qquad u_y=20\sin30^\circ=10.0\,\text{m s}^{-1}.$$

At landing, the vertical displacement is zero:

$$0=10.0t-\frac{1}{2}(9.8)t^2.$$

Ignoring $t=0$, the flight time is $t=2.04\,\text{s}$. Therefore,

$$x=u_xt=(17.3)(2.04)=35.3\,\text{m}.$$

For different launch and landing heights, do not use symmetry; determine time from vertical displacement first, then substitute it into the horizontal equation.

A common misconception is that acceleration is zero at maximum height. Only vertical velocity is momentarily zero; acceleration remains $9.8\,\text{m s}^{-2}$ downward.

Exam Technique

Draw and label both velocity components, state your sign convention, and separate horizontal and vertical working. For Calculate, show the equation, substitution, and answer. Remember that only horizontal velocity remains constant in this model.