Using suvat

• Suvat equations are a set of five kinematic formulae that relate displacement ($s$), initial velocity ($u$), final velocity ($v$), constant acceleration ($a$), and time ($t$). They are valid only when acceleration is uniform (constant), which is exactly the case in projectile motion under standard modelling assumptions (no air resistance, gravity only).
• The five equations are:

$v = u + at$ $s = ut + \frac{1}{2}at^2$ $s = vt - \frac{1}{2}at^2$ $s = \frac{1}{2}(u + v)t$ $v^2 = u^2 + 2as$

Each equation omits one of the five variables, so you choose the equation that contains the three knowns and the one unknown you need to find.

• Projectile motion is 2D, so the suvat variables $s$, $u$, $v$, and $a$ become vectors with horizontal ($x$) and vertical ($y$) components. However, because the two directions are independent of each other, you apply the 1D suvat equations separately to each component direction. The only parameter shared between the two directions is time $t$, which is the same for both.

• Key modelling assumptions underpin the use of suvat in projectile problems:

• No air resistance, so there are no horizontal forces acting on the projectile

• The only force is gravity, acting vertically downward

• Spin and rotation of the projectile are ignored

• The resulting path is a parabola, and the motion is symmetrical about the highest point (when launch and landing heights are equal)

• The fundamental reason suvat equations apply to projectile motion is that the acceleration is constant in both directions. Horizontally, the acceleration is zero ($a_x = 0$) because no horizontal forces act on the projectile (air resistance is ignored). Vertically, the acceleration is $g$ (gravitational acceleration, approximately $9.8 \text{ m s}^{-2}$) directed downward. Since both accelerations are constant, each direction independently satisfies the conditions for using suvat.

• The independence of horizontal and vertical motion is the crucial physical principle. What happens in the $x$-direction has no effect on the $y$-direction and vice versa. This means you can solve for the horizontal displacement using only horizontal quantities, and separately solve for the vertical displacement using only vertical quantities. The time $t$ is the bridge: it is the same in both directions and is often the variable you solve for first.

• Because $a_x = 0$, the horizontal suvat equations simplify dramatically. The equation $s = ut + \frac{1}{2}at^2$ becomes simply $x = (U\cos\theta)\,t$ — i.e., displacement equals speed times time. The horizontal speed never changes throughout the motion. This simplification means that horizontal problems are essentially just "distance = speed $\times$ time" calculations.

• The formula $v^2 = u^2 + 2as$ is sometimes called the "timeless" suvat equation because it does not involve $t$. It is particularly useful for finding the maximum height (where $v_y = 0$) directly without needing to find the time first. However, note that this equation is a scalar equation and must be applied to each direction separately — it does not work as a single 2D vector equation.

• Step 1 — Draw a diagram. Sketch the trajectory and clearly label the initial velocity components $u_x = U\cos\theta$ and $u_y = U\sin\theta$, the acceleration ($a_x = 0$, $a_y = \pm g$), and establish which direction is positive. This diagram prevents sign errors and helps you identify which quantities are known.

• Step 2 — Choose a positive direction and be consistent. If you take upward as positive, then $a_y = -g$ and downward displacements are negative. If you take downward as positive, then $a_y = +g$ but upward initial velocities become negative. Either convention works, but mixing them within the same equation will produce wrong answers. Write your sign convention clearly at the start.

• Step 3 — List known and unknown quantities for EACH direction separately. Write out a table:

This systematic listing ensures you identify which suvat equation to use.

• Step 4 — Select the appropriate suvat equation. Choose the equation that contains your three known quantities and the one unknown you seek. For maximum height problems, use $v^2 = u^2 + 2as$ vertically with $v_y = 0$. For time-of-flight problems, use $s = ut + \frac{1}{2}at^2$ vertically with $y = 0$ (same height landing) or $y = -h$ (different height landing). For range problems, find $t$ first from the vertical equation, then substitute into $x = (U\cos\theta)\,t$.

• Step 5 — Solve and interpret. After solving the algebra, check the sign and magnitude of your answer. A negative time should be rejected (it represents the moment before launch). A displacement can be negative if the projectile ends below its starting point. Always give your final answer to the required degree of accuracy (typically 3 significant figures).

• The following table summarises the critical differences between the horizontal and vertical analysis:

• $v = u + at$ vs $v^2 = u^2 + 2as$: Use $v = u + at$ when you know (or want) the time. Use $v^2 = u^2 + 2as$ when time is not given and not needed — for instance, finding the maximum height directly from the initial vertical speed. The second equation is a scalar equation linking speed, acceleration, and displacement without time.

• Same-level vs different-level problems: When the projectile launches and lands at the same height, the vertical displacement over the full flight is zero ($y = 0$), and the motion is symmetric — time to reach maximum height equals time to descend. When the launch and landing heights differ (e.g., launching from an elevated position), the vertical displacement $y$ is nonzero (often negative if the landing point is below the launch point), and you typically solve a quadratic equation in $t$.

• Speed vs velocity: Speed is the magnitude of the velocity vector and is always positive. At any point in the trajectory, the speed is found using Pythagoras: $\text{speed} = \sqrt{v_x^2 + v_y^2}$. The minimum speed occurs at the maximum height, where $v_y = 0$, so the minimum speed equals $|v_x| = U\cos\theta$. The speed is never zero during flight because the horizontal component is always nonzero.

• Always draw and annotate a diagram before writing any equations. Mark the initial velocity components, the acceleration directions, and your positive direction convention. Examiners look for this, and it is an easy way to pick up method marks even if the final answer has an arithmetic slip.

• Use exact values during intermediate calculations and only round at the very end. If you find a time of $t = 2.0382\ldots$ from one part, substitute the full unrounded value into the next part. Premature rounding is one of the most common causes of lost accuracy marks. Store intermediate results in your calculator memory.

• When solving a quadratic for time, you will typically get two solutions — one positive and one negative. Always reject the negative root because time cannot be negative in this context. Briefly state your reason (e.g., "reject $t < 0$ as time must be positive") to show the examiner you understand the physical interpretation.

• Check your sign convention is consistent throughout the solution. A very common error is to take upward as positive for velocity but forget to make $a = -g$ (or vice versa). If you find yourself getting a negative maximum height or a negative range for a standard problem, this is almost certainly a sign error. Go back and check every term.

• For finding speed at a given instant, remember you need both components of velocity at that time. Find $v_x$ (which is constant = $U\cos\theta$) and $v_y$ (using $v = u + at$ vertically), then compute $\text{speed} = \sqrt{v_x^2 + v_y^2}$. A common mistake is to report only $v_y$ as the speed, forgetting the horizontal component.

• Pitfall: Confusing displacement with distance. In projectile motion, the vertical displacement from start to end can be zero (same-level landing) or negative (landing below launch point), but the distance travelled along the curved path is always positive and much larger. When a problem asks for "the distance below the launch point," you need the magnitude of the vertical displacement, not the displacement itself.

• Pitfall: Applying $v^2 = u^2 + 2as$ as a 2D equation. This equation is only valid in one dimension at a time. You cannot substitute the full 2D speed for $v$ and $u$ and the full 2D displacement for $s$ in a single application. You must apply it separately: $v_x^2 = u_x^2 + 2a_x s_x$ (which just says $v_x = u_x$ since $a_x = 0$) and $v_y^2 = u_y^2 + 2a_y s_y$.

• Pitfall: Thinking the projectile stops at the maximum height. At the highest point, the vertical velocity is zero, but the horizontal velocity is still $U\cos\theta$. The projectile does not stop — it is moving horizontally at its fastest relative to the vertical (which is zero). The overall speed is at its minimum at the top, not zero.

• Pitfall: Wrong sign for vertical displacement in different-level problems. If a projectile is launched from a height $h$ above the ground and lands on the ground, the vertical displacement from launch to landing is $-h$ (if upward is positive), not $+h$. Getting this sign wrong leads to an incorrect quadratic equation and therefore an incorrect time of flight.

• Pitfall: Forgetting that the angle can be below the horizontal. Not all projectiles are launched upward. If a projectile is launched at an angle $\theta$ below the horizontal, then $u_y = -U\sin\theta$ (negative if upward is positive). Alternatively, you can treat $\theta$ as a negative angle. Failing to account for this changes the entire solution.

• Connection to the trajectory equation: By eliminating time $t$ from the horizontal and vertical suvat equations, you obtain the Cartesian trajectory equation $y = x\tan\theta - \frac{gx^2}{2U^2\cos^2\theta}$. This is a quadratic in $x$, confirming the parabolic shape. The trajectory equation is useful when you want to find the height at a given horizontal distance without needing to find the time.

• Derived projectile formulae: Applying suvat systematically yields standard results for same-level projectiles:

• Time of flight: $T = \frac{2U\sin\theta}{g}$ (set $y = 0$ in the vertical displacement equation)

• Range: $R = \frac{U^2 \sin 2\theta}{g}$ (substitute $T$ into horizontal displacement)

• Maximum height: $H = \frac{U^2 \sin^2\theta}{2g}$ (use $v_y^2 = u_y^2 + 2a_y s_y$ with $v_y = 0$)

• Time to max height: $t_H = \frac{U\sin\theta}{g}$ (use $v_y = u_y + a_y t$ with $v_y = 0$)

These only apply when launch and landing are at the same height.

• Connection to vectors and trigonometry: Resolving the initial velocity into components uses basic trigonometry ($\cos$ and $\sin$). Reconstructing the velocity vector at any time requires combining horizontal and vertical components using Pythagoras' theorem for magnitude and $\tan^{-1}$ for direction. These are the same vector skills used throughout mechanics.

• Extension to more advanced problems: Harder projectile problems may involve simultaneous equations (e.g., two projectiles meeting), using the trajectory equation to determine whether a projectile clears a given obstacle, or working backward from landing data to find the launch speed or angle. All of these ultimately rest on the same suvat framework applied to each direction independently.