Projectile motion: parametric equations, range, maximum height and time of flight

What this dot point is asking

NESA wants you to model a projectile launched from ground level (or some height) with a given initial speed and angle, write down the parametric position equations, find the trajectory equation, and compute the standard quantities: range, time of flight, maximum height, and velocity at a given time.

The whole topic rests on a single physical idea: horizontal and vertical motion are independent. Gravity acts only downward, so it changes the vertical velocity and leaves the horizontal velocity untouched. That is why a problem in two dimensions splits cleanly into two one-dimensional problems that share the same clock $t$.

The answer

The model

A projectile launched from the origin with initial speed $V$ at angle $\alpha$ above the horizontal, with gravitational acceleration $g$ downward and no air resistance:

Velocity components:

Position equations:

Acceleration:

Horizontal motion is at constant velocity; vertical motion is under constant downward acceleration $g$.

Building the trajectory, stage by stage

The figures below build a launch from the ground one feature at a time, in the order you should set the problem up.

Stage 1, resolve the launch velocity. The initial velocity $V$ at angle $\alpha$ splits into a horizontal part $V\cos\alpha$ and a vertical part $V\sin\alpha$. Every later formula is built from these two components, so resolve first.

Stage 2, trace the path. Constant horizontal velocity carries the projectile sideways at a steady rate while gravity bends the vertical motion, and the combination is a parabola. Eliminating $t$ from the two position equations gives its Cartesian form, $y = x\tan\alpha - \dfrac{gx^2}{2V^2\cos^2\alpha}$.

Stage 3, find the maximum height. At the top of the path the projectile is momentarily moving horizontally, so the vertical velocity is zero: $\dot y = 0$. Solving $V\sin\alpha - gt = 0$ gives $t = \dfrac{V\sin\alpha}{g}$; substituting into $y$ gives $H = \dfrac{V^2\sin^2\alpha}{2g}$.

Stage 4, find the range and time of flight. On level ground the projectile lands when $y = 0$ again, at $t = \dfrac{2V\sin\alpha}{g}$ - exactly twice the time to the top, by symmetry. The range is the horizontal distance covered in that time, $R = V\cos\alpha\cdot t = \dfrac{V^2\sin 2\alpha}{g}$.

Standard derived quantities

Time of flight (on horizontal ground): set $y(t) = 0$.

Range on horizontal ground: $R = V \cos \alpha \cdot T$.

(Maximum range occurs at $\alpha = 45^\circ$, because $\sin 2\alpha$ peaks there.)

Maximum height: set $\dot y(t) = 0$, giving $t = \frac{V \sin \alpha}{g}$, then substitute.

Trajectory (Cartesian) equation

Eliminate $t$ from the position equations. From $x = V \cos \alpha \cdot t$, $t = \frac{x}{V \cos \alpha}$. Substitute into $y$:

This is a downward parabola. The second form, using $\sec^2\alpha = 1 + \tan^2\alpha$, is handy when a question gives or asks for the launch angle through $\tan\alpha$.

Launched from a height

If the projectile starts at height $h$, then $y(t) = h + V \sin \alpha \cdot t - \tfrac{1}{2} g t^2$. The time of flight is now the larger root of this quadratic in $t$, and the symmetric range formula no longer applies - you must solve the quadratic. This is the single most common way the standard formulas are made to fail, so check the launch height before reaching for $R = \frac{V^2\sin 2\alpha}{g}$.

Velocity: speed and direction

Magnitude of velocity at time $t$:

Direction (above horizontal):

The speed is minimum at the top of the trajectory, where $\dot y = 0$ and the speed is just $V\cos\alpha$. On level ground the landing speed equals the launch speed, by symmetry of the parabola.

Range on inclined ground

For ground inclined at angle $\beta$ to the horizontal, the projectile lands when its path meets the line of the slope, so you set $y(t) = \pm x(t)\tan\beta$ (sign depending on uphill or downhill) and solve for $t$. This is a harder Section II application; the method is unchanged, only the landing condition is.

How exam questions ask about projectile motion

• "Show that the position is given by ..." / "Derive the equations of motion." Start from $\ddot x = 0$, $\ddot y = -g$ and integrate twice, using the initial velocity components and launch point as the constants. Show the integration; do not just quote the result.
• "Find the greatest height." A $\dot y = 0$ question: solve for $t$, substitute into $y$.
• "Find the range" or "how far from the launch point does it land?" A $y = 0$ question (level ground) or a "meets the slope / cliff base" condition; solve for $t$, then substitute into $x$.
• "Find the speed and direction after $t$ seconds / on landing." Combine the components: speed $\sqrt{\dot x^2 + \dot y^2}$, angle $\arctan(\dot y/\dot x)$. Mind the sign of $\dot y$ on the way down.
• "Show the trajectory is a parabola" or "find the Cartesian equation." Eliminate $t$ between the two position equations.
• "For what angle is the range greatest?" Maximise $\sin 2\alpha$: the answer is $45^\circ$ on level ground.
• "The projectile is launched from a height $h$ ..." Add $h$ to $y(t)$ and solve the resulting quadratic; the symmetric formulas do not apply.