Projectiles

The Core Idea, Setup Routine, and Equations of Motion

Imagine rolling a marble off the edge of a flat desk at $5\text{ m s}^{-1}$ from a height of $1.225\text{ m}$. Gravity pulls it down with acceleration $g = 9.8\text{ m s}^{-2}$, taking $t = \sqrt{\frac{2(1.225)}{9.8}} = 0.5\text{ s}$ to hit the floor. In that exact half-second, it continues moving forward at $5\text{ m s}^{-1}$, landing $5(0.5) = 2.5\text{ m}$ out from the base of the desk. The horizontal motion does not alter how fast it falls, and gravity does not alter how fast it moves sideways.

The Standard Setup Routine

Before writing down numbers or formulas on the exam, always establish your coordinate frame:

1. Sketch the path, marking the launch point, the angle of projection $\alpha$, ground level, and any cliff height or barrier.
2. Place the origin $(0,0)$ at the launch point and take up and right as positive.
3. Resolve initial velocity into perpendicular components: $u_x = u\cos\alpha$ and $u_y = u\sin\alpha$.
4. State constant accelerations: $a_x = 0$ and $a_y = -9.8\text{ m s}^{-2}$ (unless $g$ is left as a symbol).
5. Write down the four governing equations for $v_x, v_y, s_x, s_y$.

Derivation Using Calculus

By Newton's second law, $\vec{F} = m\vec{a}$. The only force in the baseline model is downward weight $-mg\vec{j}$, so $\vec{a} = 0\vec{i} - g\vec{j}$.

Integrating acceleration with respect to time gives the velocity vector:

Integrating velocity gives the displacement vector $\vec{r}(t) = s_x\vec{i} + s_y\vec{j}$:

This gives our four fundamental tools:

• $v_x = u\cos\alpha$
• $v_y = u\sin\alpha - gt$
• $s_x = (u\cos\alpha)t$
• $s_y = (u\sin\alpha)t - \frac{1}{2}gt^2$

Using the chain rule, $a_y = \frac{dv_y}{dt} = v_y\frac{dv_y}{ds_y} = -g$. Integrating with initial speed $u_y$ gives $v_y^2 = u_y^2 - 2gs_y$. This links vertical speed directly to vertical displacement without calculating time.

Vectors, Polar Form, and the Dot Product

Working from a Velocity Given as a Vector

Exam questions often specify the launch velocity directly in unit-vector form: $\vec{u} = a\vec{i} + b\vec{j}$. You do not need to resolve angles first:

• $u_x = a$ and $u_y = b$
• Launch speed $u = |\vec{u}| = \sqrt{a^2 + b^2}$
• Launch angle $\alpha = \tan^{-1}\left(\frac{b}{a}\right)$

For example, if $\vec{u} = 12\vec{i} + 20\vec{j}\text{ m s}^{-1}$, the speed is $\sqrt{12^2 + 20^2} = 23.32\text{ m s}^{-1}$ at an elevation of $\tan^{-1}\left(\frac{20}{12}\right) = 59.04^\circ$. After $t$ seconds, velocity is $\vec{v} = 12\vec{i} + (20 - 9.8t)\vec{j}$ and displacement is $\vec{r} = 12t\vec{i} + (20t - 4.9t^2)\vec{j}$. If asked for velocity at a specific point, state it in $\vec{i},\vec{j}$ format or give the speed and angle.

Direction of Motion

At any point in flight, the direction of motion makes an angle $\theta$ with the horizontal where $\tan\theta = \frac{v_y}{v_x}$. Calculate the magnitude $\theta = \tan^{-1}\left(\frac{|v_y|}{v_x}\right)$ and state clearly in words whether the particle is travelling above the horizontal (ascending, $v_y > 0$) or below the horizontal (descending, $v_y < 0$).

Perpendicular Velocity Vectors via the Dot Product

Two vectors are perpendicular if and only if their dot product is zero. This provides a direct method to find when a projectile moves at right angles to its launch direction: set $\vec{u}\cdot\vec{v} = 0$.

This instant occurs before landing only if $t < T$, meaning $\frac{u}{g\sin\alpha} < \frac{2u\sin\alpha}{g} \implies \sin^2\alpha > \frac{1}{2} \implies \alpha > 45^\circ$. If $\alpha \le 45^\circ$, the projectile strikes level ground before turning through $90^\circ$.

Standard Derivations on Level Ground

When a projectile launches and lands on the same horizontal plane, the flight path is symmetrical.

Greatest Height

The particle reaches its highest point when the vertical component of velocity momentarily drops to zero: $v_y = 0$ at greatest height.

Substitute this time into vertical displacement to derive the greatest height ($H$):

At this apex, the particle still has horizontal velocity $v_x = u\cos\alpha$.

Time of Flight

The particle lands when vertical displacement returns to zero ($s_y = 0$):

Ignoring $t = 0$ gives the total time of flight ($T$):

Horizontal Range and Maximum Range

The horizontal range ($R$) is the distance covered horizontally during flight time $T$:

Using the identity $\sin 2\alpha = 2\sin\alpha\cos\alpha$:

For a fixed speed $u$, the range is greatest when $\sin 2\alpha$ is as large as possible. The maximum value of sine is $1$, so $\sin 2\alpha = 1 \implies 2\alpha = 90^\circ \implies \alpha = 45^\circ$. The maximum range is:

Complementary angles (angles summing to $90^\circ$, like $30^\circ$ and $60^\circ$) yield identical ranges because $\sin(2(90^\circ - \alpha)) = \sin(180^\circ - 2\alpha) = \sin 2\alpha$. The lower angle gives a flatter, faster flight; the higher angle creates a steep, high lob.

Launches from Cliffs, Heights, and Depressions

Greatest Height Above the Ground Versus Rise Above the Launch Point

The formulas $H = \frac{u^2\sin^2\alpha}{2g}$ and $v_y^2 = u_y^2 - 2gs_y$ calculate the vertical rise relative to the point of projection. If a particle is launched from a cliff of height $h$ and reaches a maximum height $M$ above the ground, the rise above the launch point is $s_y = M - h$. Always subtract the platform height before calculating $u_y$ or $\alpha$.

Landing Below the Launch Point

Taking the origin at launch and up as positive, ground level situated a vertical drop $h$ below the launch point corresponds to $s_y = -h$ for a launch from a cliff.

This quadratic yields one positive root for flight time. Level-ground formulas like $\frac{u^2\sin 2\alpha}{g}$ cannot be used here.

Conservation of energy confirms the final impact speed: $\frac{1}{2}mu^2 + mgh = \frac{1}{2}mv^2 \implies v = \sqrt{u^2 + 2gh}$. This speed depends purely on $u$ and drop height $h$, completely independent of the launch angle.

Projection Below the Horizontal

When an object is thrown downwards at an angle $\alpha$ below the horizontal, keep up as positive. The only adjustment is the sign of the vertical velocity:

• $u_x = u\cos\alpha$
• $u_y = -u\sin\alpha$
• $a_y = -g$
• Target ground level remains $s_y = -h$

A horizontal launch is simply $\alpha = 0$, giving $u_x = u$ and $u_y = 0$.

The Trajectory Equation, Targets, and Obstacles

The Parabola and the Trajectory Equation

Eliminating time $t$ gives the path coordinates directly. From horizontal motion, $t = \frac{x}{u\cos\alpha}$. Substituting into vertical displacement:

Using the identity $\frac{1}{\cos^2\alpha} = \sec^2\alpha = 1 + \tan^2\alpha$, we obtain the trajectory equation:

Because this has the algebraic form $y = px - qx^2$ where $p$ and $q$ are constants with $q > 0$, the path of a projectile in free flight is a parabola opening downwards.

Target Quadratics in tan α

Rearranging the trajectory equation for a target $(X, Y)$ yields a quadratic in $\tan\alpha$:

• Two positive roots give two projection angles: a low drive (rapid flight) and a high lob.
• A negative root indicates an angle below the horizontal, valid when aiming downward.
• If the discriminant is negative, no real angle exists: the target is out of range at speed $u$.

Two Times at the Same Height and Clearing Obstacles

Setting $s_y = k$ gives $\frac{1}{2}gt^2 - (u\sin\alpha)t + k = 0$. The smaller root $t_1$ is the time on the way up; the larger root $t_2$ is on the way down. The duration spent above height $k$ is $t_2 - t_1$.

To prove a projectile clears an obstacle of height $H_{\text{obs}}$ at distance $d$:

1. Find the time to reach the barrier: $t = \frac{d}{u\cos\alpha}$.
2. Calculate $s_y$ at this time.
3. Compare: if $s_y > H_{\text{obs}}$, the projectile clears it by $s_y - H_{\text{obs}}$. State this comparison explicitly in words to secure the conclusion mark.

Mid-Air Collisions and Staggered Launches

Two projectiles collide when their spatial coordinates match at the same instant: $s_{x,A} = s_{x,B}$ and $s_{y,A} = s_{y,B}$.

Simultaneous Launch from Opposite Sides

If particles $A$ and $B$ are fired at $t = 0$ toward each other from two points on the same horizontal level, separated by distance $D$:

• Horizontal: $(u_A\cos\alpha_A)t + (u_B\cos\alpha_B)t = D$
• Vertical: $(u_A\sin\alpha_A)t - \frac{1}{2}gt^2 = (u_B\sin\alpha_B)t - \frac{1}{2}gt^2$

Cancelling $-\frac{1}{2}gt^2$ reveals that their vertical velocities must match: $u_A\sin\alpha_A = u_B\sin\alpha_B$. Both particles rise and fall in synchrony throughout flight. If the launch points are at different heights, equate absolute heights instead: $h_A + s_{y,A} = h_B + s_{y,B}$, and the gravity terms still cancel.

Staggered Launches

If projectile $B$ launches $d$ seconds after projectile $A$, and collision occurs at time $t$ (measured from $A$'s release), then particle $B$ has been airborne for only $\tau = t - d$ seconds. The gravity terms do not cancel because $-\frac{1}{2}gt^2 \ne -\frac{1}{2}g(t - d)^2$. Set up $s_{y,A}(t) = s_{y,B}(t - d)$ and solve with the given time delay.

Worked illustration: $P$ is projected at $t = 0$ from $A$ with speed $38\text{ m s}^{-1}$ at $41^\circ$ to $AB$; $Q$ is projected from $B$ one second later with speed $u$ at $64^\circ$ to $BA$; they collide at $t = 3\text{ s}$. Equating heights: $38\sin 41^\circ(3) - 4.9(3)^2 = u\sin 64^\circ(2) - 4.9(2)^2$, giving $74.81 - 44.1 = 1.7975u - 19.6$, so $u = 28.0\text{ m s}^{-1}$; then $D = 38\cos 41^\circ(3) + 28\cos 64^\circ(2) = 86.05 + 24.55 = 110.6\text{ m}$.

Mathematical Modelling in Context

Every projectile formula relies on idealised assumptions. The examination contextualises questions within Strand 1, frequently asking candidates to evaluate assumptions, consider limitations, or suggest model refinements.

Model Assumptions and Physical Effects

• Assumption: The object is a point particle. Neglecting dimensions means shape, orientation, and aerodynamic spin (such as backspin generating Magnus lift) are omitted from the equations.
• Assumption: Air resistance is negligible. In reality, fluid drag acts opposite to velocity ($F_{\text{drag}} \propto v^n$). Drag reduces horizontal velocity continuously, lowers maximum height and range, steepens the descent relative to ascent, and lowers the optimal range angle below $45^\circ$.
• Assumption: Gravitational acceleration is constant ($g = 9.8\text{ m s}^{-2}$). This holds accurately near the surface of the Earth over typical sports trajectories.

Sample Answers for the Modelling Question

• Assumptions: "I assumed the ball acts as a particle, ignoring its size and spin. I assumed air resistance is negligible, meaning gravity is the sole force acting during flight. I also assumed $g = 9.8\text{ m s}^{-2}$ is uniform."
• Effect on real performance: "Air resistance opposes motion, so a real ball travels a shorter distance and reaches a lower peak than the calculated model predicts. The model over-estimates range, and the true trajectory is asymmetric with a steeper drop."
• Model refinement: "Refine the model by introducing a resistive drag term proportional to velocity or velocity squared ($-kv$ or $-kv^2$), solving the resulting differential equations analytically or numerically."

Note on curriculum scope: Projectiles launched onto inclined planes appeared in pre-2023 examinations. The current specification focuses on horizontal planes and targets.

Key terms

Time of Flight

The total elapsed time from release until the projectile reaches ground or target level.

Greatest Height

The maximum vertical displacement attained above the launch position, occurring when vertical velocity $v_y = 0$.

Horizontal Range

The total horizontal displacement achieved over the flight time on level ground.

Maximum Range

The greatest horizontal distance achievable for a given launch speed, occurring at a $45^\circ$ angle of projection on level ground.

Angle of Projection

The angle $\alpha$ between the initial velocity vector and the horizontal reference plane.

Trajectory Equation

The cartesian relation $y = x\tan\alpha - \frac{gx^2}{2u^2}(1+\tan^2\alpha)$ defining the path in space independent of time.

Parabola

The downward-curving quadratic path $y = px - qx^2$ followed by any particle in uniform free-fall kinematics.

Dot Product

The algebraic product $\vec{u}\cdot\vec{v} = u_x v_x + u_y v_y$, which evaluates to zero when vectors are perpendicular.

Complementary Angles

Pairs of launch angles summing to $90^\circ$ that yield identical ranges on flat ground for the same initial speed.