ProjectilesAQA A-Level Maths: Revision notes

A projectile is a particle moving under gravity alone. The standard modelling assumptions are: the object is a particle (no size, no spin), there is no air resistance, the acceleration is constant at $g=9.8\ \text{m s}^{-2}$ downwards, and the motion takes place in a vertical plane. In vector form, with $\mathbf{i}$ horizontal and $\mathbf{j}$ vertically upwards, $\mathbf{a}=-g\mathbf{j}=(0,-9.8)$. The key idea is that horizontal and vertical motion are independent: horizontally there is no acceleration, vertically the acceleration is $-g$.

If a particle is projected with speed $U$ at angle $\theta$ above the horizontal, resolve the velocity: $u_x=U\cos\theta,\qquad u_y=U\sin\theta.$ Horizontally ($a_x=0$): $x=U\cos\theta\,t$, and $v_x=U\cos\theta$ always. Vertically ($a_y=-g$): $v_y=U\sin\theta-gt$, $\;y=U\sin\theta\,t-\frac12gt^2$, $\;v_y^2=U^2\sin^2\theta-2gy$. In vector form the same results are $\mathbf{v}=\mathbf{u}+\mathbf{a}t$ and $\mathbf{r}=\mathbf{u}t+\frac12\mathbf{a}t^2$ with $\mathbf{a}=-g\mathbf{j}$. For $\mathbf{u}=(6\mathbf{i}+14.7\mathbf{j})$ after $2$ s: $\mathbf{v}=6\mathbf{i}+(14.7-19.6)\mathbf{j}=6\mathbf{i}-4.9\mathbf{j}$.

For a projectile launched from and landing on the same horizontal level:

• Time to greatest height: vertical velocity is zero, so $t=\frac{U\sin\theta}{g}$.
• Greatest height: $H=\frac{U^2\sin^2\theta}{2g}$.
• Time of flight: $y=0$ gives $T=\frac{2U\sin\theta}{g}$, twice the time to the top.
• Range: $R=U\cos\theta\times T=\frac{U^2\sin2\theta}{g}$. Example: $U=20$, $\theta=30^\circ$: $u_y=10$, so $T=\frac{20}{9.8}=2.04$ s and range $=17.32\times2.04=35.3$ m. If the launch and landing heights differ, do not use these formulae: set $y$ equal to the displacement and solve the quadratic in $t$ (for example a stone thrown horizontally from $19.6$ m: $-19.6=-4.9t^2$, so $t=2$ s).

At any time the velocity has components $v_x$ and $v_y$. Then $\text{speed}=\sqrt{v_x^2+v_y^2},\qquad \tan\phi=\left|\frac{v_y}{v_x}\right|,$ where $\phi$ is the angle of the direction of motion to the horizontal. A positive $v_y$ means the particle is rising; a negative $v_y$ means it is falling. At the highest point $v_y=0$, so the velocity is horizontal and the speed is $v_x$ (the minimum speed). Example: for $\mathbf{u}=(6\mathbf{i}+14.7\mathbf{j})$, at height $5$ m $v_y^2=14.7^2-2(9.8)(5)=118.09$, so speed $=\sqrt{36+118.09}=12.4\ \text{m s}^{-1}$. The speed is the same going up or down at a given height.

Eliminating $t$ from $x=U\cos\theta\,t$ and $y=U\sin\theta\,t-\frac12gt^2$ gives the equation of the path: $y=x\tan\theta-\frac{gx^2}{2U^2\cos^2\theta}.$ This is a parabola. For $U=25$, $\tan\theta=\frac34$ it is $y=0.75x-0.01225x^2$, which can be used to test whether the particle clears an obstacle at a given $x$, or to find where it is at a given height by solving a quadratic. Typical tasks: find when and where a particle lands, whether it clears a wall or hits a target, the speed and direction of impact, and the effect of modelling assumptions (air resistance would reduce range and height; wind and spin would alter the path).