Projectile motionEdexcel A-Level Maths: Revision notes

A projectile is a body moving under gravity alone. At A Level you model it as a particle (no size, so no spin or rotation) moving in a vertical plane, with no air resistance and a constant gravitational acceleration $g=9.8$ m s$^{-2}$ (unless the question gives another value such as $10$) acting vertically downwards. Take $\mathbf{i}$ horizontal, $\mathbf{j}$ vertically upwards, so $\mathbf{a}=-g\mathbf{j}$. The horizontal and vertical motions are independent: horizontally there is no acceleration; vertically the acceleration is $-g$. If a projectile is launched with speed $U$ at angle $\alpha$ above the horizontal, then $\mathbf{u}=U\cos\alpha\,\mathbf{i}+U\sin\alpha\,\mathbf{j}$.

$\text{Horizontal: } x=Ut\cos\alpha,\qquad \text{Vertical: } y=Ut\sin\alpha-\tfrac12gt^2.$ Velocity components: $v_x=U\cos\alpha$ (constant) and $v_y=U\sin\alpha-gt$. In vector form, $\mathbf{r}=(U\cos\alpha\,t)\mathbf{i}+(U\sin\alpha\,t-\tfrac12gt^2)\mathbf{j}$. The speed at any time is $\sqrt{v_x^2+v_y^2}$, and the direction of motion makes an angle $\tan^{-1}\left(\frac{|v_y|}{v_x}\right)$ with the horizontal, above it if $v_y>0$ and below it if $v_y<0$. Worked example: thrown horizontally at $12$ m s$^{-1}$ from a $45$ m cliff. Vertically $45=\frac12(9.8)t^2$, so $t=3.03$ s; the horizontal distance is $12\times3.03=36.4$ m; $v_y^2=2(9.8)(45)=882$ so the impact speed is $\sqrt{144+882}=32.0$ m s$^{-1}$.

For a projectile that starts and ends at the same horizontal level:

• Time of flight: $y=0$ gives $T=\dfrac{2U\sin\alpha}{g}$.
• Greatest height: at the top $v_y=0$, which occurs at $t=\frac{U\sin\alpha}{g}$ (half the flight). Then $H=\dfrac{U^2\sin^2\alpha}{2g}$.
• Range: $R=U\cos\alpha\times T=\dfrac{2U^2\sin\alpha\cos\alpha}{g}=\dfrac{U^2\sin2\alpha}{g}$. For a given $U$ the range is greatest when $\sin2\alpha=1$, that is $\alpha=45^\circ$, and then $R_{\max}=\frac{U^2}{g}$. Example: $U=20$, $\alpha=30^\circ$: $T=2.04$ s, $H=5.10$ m, $R=35.3$ m.

Questions may ask you to show the results above, so be able to derive them.

1. Write $y=Ut\sin\alpha-\frac12gt^2$ and set $y=0$: $t\left(U\sin\alpha-\frac12gt\right)=0$. Reject $t=0$ to get $T=\frac{2U\sin\alpha}{g}$.
2. Substitute $t=T$ into $x=Ut\cos\alpha$ and use $\sin2\alpha=2\sin\alpha\cos\alpha$ to get $R$.
3. Set $v_y=0$ for the top, or use $v_y^2=U^2\sin^2\alpha-2gH=0$, to get $H$.

Eliminate $t$ using $t=\frac{x}{U\cos\alpha}$: $y=x\tan\alpha-\frac{gx^2}{2U^2\cos^2\alpha}.$ This is a downward-facing parabola through $O$, which is why the trajectory is a parabola. Use it to find the height at a given horizontal distance, or the horizontal distances at a given height. Example: $U=14$, $\tan\alpha=\frac34$ ($\cos^2\alpha=\frac{16}{25}$): $y=\frac34x-\frac{5}{128}x^2$. Setting $y=2$ gives $5x^2-96x+256=0$, so $x=3.2$ or $x=16$ (the ball is $2$ m high on the way up and on the way down).

Typical problems: does the ball clear a wall or pole at a given distance? what is the speed and direction at impact? what launch angle gives a particular range? Compare the height at the wall's horizontal distance with the wall's height, and state the conclusion. The model is a simplification. Ignoring air resistance overestimates the range and greatest height; treating the object as a particle ignores spin and its size; assuming constant $g$ ignores the small variation of $g$ with height and location; and a real wind would add a horizontal force. To improve the model, include air resistance, spin or wind.