Motion in a vertical plane and projectilesEdexcel International A Level Further Maths: Revision notes

A projectile is modelled as a particle moving freely under gravity. The standard assumptions are:

• air resistance is negligible;
• the only force is weight, so the acceleration is constant, $g=9.8\text{ m s}^{-2}$ downwards;
• the size, spin and shape of the object do not matter;
• $g$ is the same at every height reached. The motion in the vertical and horizontal directions can be treated independently, linked only by the time $t$.

For motion in a vertical line use the constant-acceleration equations with $a=-g$ (upwards positive): $v=u-gt,\quad s=ut-\tfrac12gt^2,\quad v^2=u^2-2gs,\quad s=\tfrac12(u+v)t.$ At the highest point $v=0$. Going up and coming back down to the same level takes equal times, and the speeds at that level are equal. If the particle starts above the ground, $s$ is negative when it reaches the ground, giving a quadratic in $t$: take the positive root. Example: thrown up at $14\text{ m s}^{-1}$ from $2$ m high. At the top, $s=\frac{14^2}{2(9.8)}=10$ m, so the greatest height above the ground is $12$ m.

For a particle projected at speed $u$ at angle $\alpha$ above the horizontal, resolve into components: $u_x=u\cos\alpha,\qquad u_y=u\sin\alpha.$ Horizontally there is no acceleration, so the velocity stays $u\cos\alpha$ and $x=u\cos\alpha\,t$. Vertically the acceleration is $-g$, so $v_y=u\sin\alpha-gt$ and $y=u\sin\alpha\,t-\tfrac12gt^2$. If you are given $\sin\alpha=\frac35$ or $\tan\alpha=\frac43$, draw a right-angled triangle to get exact values of $\cos\alpha$ and $\sin\alpha$.

For projection from level ground with speed $u$ at angle $\alpha$:

• time of flight: $y=0$ gives $T=\dfrac{2u\sin\alpha}{g}$;
• greatest height: $v_y=0$ gives $H=\dfrac{u^2\sin^2\alpha}{2g}$;
• range: $R=u\cos\alpha\times T=\dfrac{u^2\sin2\alpha}{g}$. Derive these from the component equations rather than memorising them; they apply only when the landing point is at the same level as the launch point. Example: $u=28$, $\sin\alpha=\frac35$ gives $T=\frac{24}{7}$ s, $H=14.4$ m and $R=76.8$ m.

At any time the velocity has components $v_x=u\cos\alpha$ and $v_y=u\sin\alpha-gt$. The speed is $\sqrt{v_x^2+v_y^2}$ and the angle $\phi$ to the horizontal satisfies $\tan\phi=\dfrac{|v_y|}{|v_x|}$; the particle is moving upwards if $v_y>0$ and downwards if $v_y<0$. When a particle is projected from a height, or has to clear an obstacle, use the time found from the horizontal equation, then substitute into the vertical equation. If it lands below the launch level, put the negative displacement into $y$ and solve the quadratic; reject the negative root. Example: from a $40$ m cliff, $u=20$ at $30^\circ$: $-40=10t-4.9t^2$ gives $t=4.05$ s, and the speed on landing is $\sqrt{20^2+2(9.8)(40)}=34.4\text{ m s}^{-1}$.