Resolve projectile motion into independent horizontal (constant velocity) and vertical (constant acceleration) components to predict range, time of flight and maximum height.

What this dot point is asking

You need to model the flight of a projectile by resolving the launch velocity into horizontal and vertical components, treating each component independently, and using the shared time of flight to connect them. From this you can predict time of flight, range, maximum height and the velocity at any instant.

The key idea: independence of the two directions

Once a projectile leaves the hand, ramp or muzzle, the only force on it is gravity (we ignore air resistance at this level). Gravity acts vertically, so:

• Horizontal: no force, so no acceleration. The horizontal velocity $u_x$ stays constant for the whole flight.
• Vertical: constant acceleration $g$ downward, so the vertical velocity changes uniformly.

The two motions share one thing only: time. The projectile spends the same time $t$ in the air for both its horizontal and vertical travel. This is what lets you solve one direction for $t$ and feed it into the other.

The equations you use

Horizontal (constant velocity):

Vertical (constant acceleration, taking up as positive, $a = -g$):

Be consistent with your sign convention. Many students take "up positive" so $g$ enters as $-9.8 \text{ m s}^{-2}$; others take "down positive" for a falling object. Either works as long as you keep it consistent across the whole problem.

Maximum height, time of flight and range

At maximum height the vertical velocity is momentarily zero ($v_y = 0$), but the horizontal velocity is unchanged. Setting $v_y = 0$ gives the time to the top, and doubling it (for a projectile landing at launch height) gives the total time of flight.

• Time to peak: $t_{\text{up}} = u_y / g$
• Maximum height above launch: $H = u_y^2 / (2g)$
• Range (level ground): $R = u_x \times t_{\text{flight}}$

Velocity at any point

The velocity at any instant is the vector sum of the (constant) horizontal component and the (changing) vertical component:

where $\alpha$ is the angle of the velocity below or above the horizontal. The speed is least at the top of the path (where $v_y = 0$ and $v = v_x$).

The trajectory is a parabola

Eliminating $t$ between $x = u_x t$ and $y = u_y t - \tfrac{1}{2}gt^2$ gives $y$ as a quadratic in $x$ - a parabola. This is why projectile paths are symmetric about the highest point when launch and landing heights are equal.