3.1 - The equation of a trajectory | Cambridge A-Level Further Mathematics

The path of a projectile can be described without referring to time. You will derive its Cartesian equation, recover unknown launch information from points on the path, and decide which possible launches reach a target while clearing an obstacle.

Eliminating time

Can you tell whether a projectile passes through a point without first finding when it gets there? Suppose its motion from the origin is $x=12t$ , $y=16t-5t^2$ , with metres and seconds as units. Since $t=x/12$ , substitution gives

y=\frac43x-\frac{5x^2}{144}.

This is the path as a relation between coordinates: its Cartesian trajectory equation.

The horizontal equation is especially useful because its velocity is constant. For a launch at speed $u$ and angle $\theta$ , with $u\cos\theta\ne0$ ,

\begin{aligned} x&=u\cos\theta\,t \\ t&=\frac{x}{u\cos\theta} \\ y&=u\sin\theta\left(\frac{x}{u\cos\theta}\right) -\frac g2\left(\frac{x}{u\cos\theta}\right)^2 \end{aligned}

Simplify the first term using $\sin\theta/\cos\theta=\tan\theta$ . In the second term, the whole time expression is squared, including its denominator.

Cartesian trajectory

With origin at launch, horizontal $x$ , upward $y$ , and nonzero horizontal launch velocity,

y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}.

Equivalently, replace $1/\cos^2\theta$ by $\sec^2\theta$ . This equation is in MF19, using $V$ for launch speed. You must also be able to derive it from the component equations.

For a rightward launch use $u\cos\theta>0$ and the portion $x\ge0$ before first impact. The complete parabola includes points that may lie outside the physical flight. A vertical launch has $x=0$ throughout and cannot be described by dividing by its horizontal velocity.

Worked example 1: deriving a numerical path

Question. A particle is projected from $O$ with horizontal and upward velocity components $12$ and $16\,\mathrm{m\,s^{-1}}$ . Derive its trajectory equation and find its height when $x=24\,\mathrm m$ , before impact. Use $g=10\,\mathrm{m\,s^{-2}}$ .

Plan. Write both displacement equations and eliminate their common time.

\begin{aligned} x&=12t & y&=16t-5t^2 \\ t&=\frac{x}{12} & y&=\frac43x-\frac{5x^2}{144} \end{aligned}

At $x=24$ ,

y=\frac43(24)-\frac{5(24)^2}{144}=32-20=12.

Answer. The trajectory is $y=4x/3-5x^2/144$ , and the height is $12\,\mathrm m$ above $O$ .

Check. Horizontal motion gives $t=24/12=2$ . Substituting into the original vertical motion gives $16(2)-5(2)^2=12$ , independently confirming the point.

Change only the vertical datum. A height offset must survive the elimination of time.

Worked example 2: keeping a ground-based height

Question. The same launch takes place from a platform $6\,\mathrm m$ above ground. Let $Y$ be height above ground and $x$ the horizontal displacement from launch. Derive the trajectory and find $Y$ at $x=24\,\mathrm m$ .

Plan. Include the launch height in the vertical equation before eliminating time.

\begin{aligned} x&=12t & Y&=6+16t-5t^2 \\ Y&=6+\frac43x-\frac{5x^2}{144} \end{aligned}

Answer. The trajectory is $Y=6+4x/3-5x^2/144$ , giving $Y=18\,\mathrm m$ at $x=24\,\mathrm m$ .

Check. At $x=0$ , the equation gives $Y=6$ , the stated launch height. Subtracting the platform height from $18$ gives the $12\,\mathrm m$ rise in the preceding example.

The trajectory equation predicts coordinates, while the component equations still give time and velocity. Keep both descriptions available.

Watch out

The wrong substitution $t^2=x^2/(u\cos\theta)$ misses a square in the denominator. Correctly, $t^2=x^2/(u^2\cos^2\theta)$ .

Eliminate time in a fresh component model.

Unknown speed or angle

A point on the trajectory gives an equation connecting the launch speed and angle, so supplied information about one can determine the other.

With a known angle, substitute the target coordinates and solve for $u^2$ , remembering that speed is positive. With known speed and unknown angle, use

\sec^2\theta=1+\tan^2\theta.

Writing $p=\tan\theta$ turns the trajectory equation into a quadratic in $p$ . This is often simpler than trying inverse trigonometric functions before the algebra is finished.

Worked example 3: finding an unknown speed

Question. A particle is projected at $45^\circ$ above the horizontal from $O$ and passes through $(30,10)$ , with coordinates in metres. Find its initial speed. Use $g=10\,\mathrm{m\,s^{-2}}$ .

Plan. Put the point and known angle into the trajectory equation, then solve for the positive speed.

Since $\tan45^\circ=1$ and $\cos^245^\circ=1/2$ ,

\begin{aligned} 10&=30-\frac{10(30)^2}{2u^2(1/2)} \\ 20&=\frac{9000}{u^2} \\ u^2&=450 \end{aligned}

Answer. $u=15\sqrt2\,\mathrm{m\,s^{-1}}$ , or $21.2\,\mathrm{m\,s^{-1}}$ .

Check. Its components are both $15\,\mathrm{m\,s^{-1}}$ . It reaches $x=30$ at $t=2$ , when $y=15(2)-5(2)^2=10$ . The positive time confirms that the point is reached after launch.

Now keep the point but supply the speed instead of the angle. A second launch may satisfy exactly the same target condition.

Worked example 4: finding two possible angles

Question. A particle is projected at speed $15\sqrt2\,\mathrm{m\,s^{-1}}$ and passes through $(30,10)$ . Find the possible angles $\theta$ , where $0<\theta<90^\circ$ , using $g=10\,\mathrm{m\,s^{-2}}$ .

Plan. Substitute the point, put $p=\tan\theta$ , and solve the resulting quadratic.

\begin{aligned} 10&=30p-\frac{10(30)^2}{2(450)}(1+p^2) \\ 10&=30p-10(1+p^2) \\ p^2-3p+2&=0 \\ (p-1)(p-2)&=0 \end{aligned}

Both positive roots correspond to angles in the stated interval.

Answer. $\theta=45^\circ$ or $\theta=\tan^{-1}2=63.4^\circ$ to 1 decimal place.

Check. The first launch was checked above. For $\tan\theta=2$ , the components are $u_x=3\sqrt{10}$ and $u_y=6\sqrt{10}$ , so the target time is $30/(3\sqrt{10})=\sqrt{10}$ . Then $y=6\sqrt{10}(\sqrt{10})-5(\sqrt{10})^2=60-50=10$ .

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3.1 - The equation of a trajectory

Eliminating time

Unknown speed or angle

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