Projectile motion: height, flight time and range

In projectile problems, the horizontal and vertical equations describe the same journey. The useful connection is time: find it from one direction, then use it in the other. Here is a method and a worked example.

The connection between the two directions

A projectile is anything moving under the influence of gravity alone (ignoring air resistance). The key insight? Horizontal and vertical motion are independent.

This means:

• Horizontal velocity stays constant
• Vertical velocity changes due to gravity
• You can treat them separately!

Five steps to set up the problem

Step 1: List what you know

Before diving into equations, make a clear table:

Quantity	Horizontal	Vertical
Initial velocity	$u_x = ?$	$u_y = ?$
Final velocity	$v_x = ?$	$v_y = ?$
Acceleration	$0$	$-9.8\text{ m/s}^2$
Displacement	$s_x = ?$	$s_y = ?$
Time	$t$ (same for both)	$t$

The tip: “What are you trying to find?” should be your first question. This prevents you from using the wrong equations.

Step 2: Resolve the initial velocity

This is where many students stumble. If a ball is kicked at 15 m/s at 30° to the horizontal, you need:

$$u_x = 15 \cos(30°) = 13.0 \text{ m/s}$$ $$u_y = 15 \sin(30°) = 7.5 \text{ m/s}$$

Common mistake: Using the wrong angle or forgetting to convert to m/s.

Step 3: Choose where to start

Typically, horizontal motion is simpler - just $s_x = u_x t$ because there’s no acceleration.

My heuristic: If the problem asks “how far?” or “how long?”, start with horizontal if you know the time, or vertical if you need to find the time.

Step 4: Use the same time

Time is the same for both motions! Once you know it from one direction, use it for the other.

Step 5: Choose the right condition

Finding maximum height: Use $v_y^2 = u_y^2 + 2a_y s_y$ in the vertical direction with $v_y = 0$ at the top.

Finding range: Find the time of flight from the vertical displacement at landing, then use that time with horizontal motion. For landing at the launch height, $s_y = 0$.

Finding velocity at a point: Find horizontal and vertical velocities separately, then use Pythagoras for the magnitude.

Worked example: a football kick

A worked example of the method, assuming no air resistance and landing at the launch height.

Problem: A football is kicked with an initial velocity of 20 m/s at 35° to the horizontal. Find: a) The maximum height b) The time of flight c) The range

Working:

Part (a) - Maximum height

$$u_y = 20 \sin(35°) \approx 11.47 \text{ m/s}$$

At max height, $v_y = 0$. Using $v^2 = u^2 + 2as$:

$$0^2 = (20\sin(35°))^2 + 2(-9.8)s_y$$ $$s_y = \frac{(20\sin(35°))^2}{2 \times 9.8} \approx 6.71 \text{ m}$$

Part (b) - Time of flight

When it lands, $s_y = 0$. Using $s = ut + \frac{1}{2}at^2$:

$$0 = (20\sin(35°))t - 4.9t^2$$ $$t(20\sin(35°) - 4.9t) = 0$$

Therefore: $t = 0$ (start) or $t \approx 2.34$ s (landing)

Time of flight $\approx 2.34$ s. Keep unrounded values until the final answer.

Part (c) - Range

$$u_x = 20 \cos(35°) \approx 16.38 \text{ m/s}$$ $$s_x = (20\cos(35°))\frac{2(20\sin(35°))}{9.8} \approx 38.4 \text{ m}$$

Common mistakes to check

1. Mixing sign conventions — with upwards chosen as positive, gravitational acceleration is $-9.8$ m/s². Keep the same convention throughout.

2. Not completing the calculation - finding the projectile equation but forgetting to substitute the time back in.

3. Angle confusion - make sure you’re using the angle to the horizontal, not the vertical.

4. Mixing up displacement and distance - these are different! Displacement can be negative.

Useful results and their assumptions

For an ideal projectile with no air resistance, taking launch height as zero, you can derive:

• Maximum height is always $\frac{u_y^2}{2g}$
• Time of flight is $\frac{2u_y}{g}$ when it lands at the launch height
• Range is $\frac{u^2 \sin(2\theta)}{g}$ when it lands at the launch height

But don’t just memorize these - understand WHY they work!

When to use SUVAT or a trajectory equation

SUVAT: Use when you have specific points to analyze

• “Find velocity after 2 seconds”
• “How far does it go in 3 seconds?”

Projectile equations: Use when you need the general trajectory

• “Find the equation of the path”
• “Prove this is a parabola”

Often, you’ll use SUVAT to find specific values, then use those in projectile equations.

A useful limiting case

The biggest breakthrough comes when students realize that “straight up and down” is just a special case of projectile motion where $u_x = 0$. All the same principles apply!

What to practise next

Start with simple problems (level ground, no obstacles) before moving to more complex scenarios. Practice resolving velocities until it’s second nature - this skill carries over to circular motion, forces, and beyond!