Two launch angles for an approaching target · source reference

Printed question

The diagram shows the trajectory of a cannonball fired off from an origin $O$ with an initial speed of $v\,\mathrm{m\,s^{-1}}$ and at an angle of $\theta^\circ$ above the ground. At time $t$ seconds, its position can be modelled by $x=(v\cos\theta)t$, $y=(v\sin\theta)t-5t^2$, where $x$ m is horizontal distance with respect to $O$ and $y$ m is vertical distance with respect to ground level.

(i) Find the horizontal distance, $d$ m, that a cannonball would have travelled by the time it hits the ground. Leave your answer in terms of $v$ and $\theta$. [4]

Use $v=200$ to answer the remaining parts. An approaching target is travelling at a constant speed of $10\,\mathrm{m\,s^{-1}}$ along the ground. A cannonball is fired towards it when it is 3000 m away. You may assume its height is negligible.

(ii) Show that in order to hit the target, the possible angles at which the cannonball should be fired are $22.7^\circ$ and $69.5^\circ$. [2]

(iii) Explain at which angle the cannonball should be fired in order to hit the target earlier. [2]

(iv) Given $\theta=22.7$, find the angle that the tangent to the trajectory makes with the horizontal when $x=370$. [4]

Printed working

(i) To determine the range consider $y=0$: $(v\sin\theta)t-5t^2=0\Rightarrow t[v\sin\theta-5t]=0$.

$t=0$ is rejected as the starting point; otherwise $v\sin\theta-5t=0\Rightarrow t=v\sin\theta/5$.

At this time $x=d$, so $d=(v\cos\theta)(v\sin\theta/5)=v^2\sin\theta\cos\theta/5$.

(ii) Time for cannonball to hit ground equals time for target to reach impact:

$$\frac{v\sin\theta}{5}=\frac{3000-d}{10}\Rightarrow2v\sin\theta=3000-\frac{v^2\sin\theta\cos\theta}{5}.$$

Given $v=200$, $(200)^2\sin\theta\cos\theta/5+400\sin\theta=3000$. GC gives $22.7^\circ$ or $69.5^\circ$ (1 d.p.; shown).

(iii) Since $t=v\sin\theta/5$ and $v\sin22.7^\circ/5<v\sin69.5^\circ/5$ for $v>0$, fire at $22.7^\circ$ to hit earlier.

(iv) $x=(200\cos22.7^\circ)t$, $y=(200\sin22.7^\circ)t-5t^2$. Thus $dx/dt=184.51$, $dy/dt=77.181-10t$, $dy/dx=(77.181-10t)/184.51$.

When $x=370$, $184.51t=370\Rightarrow t=2.0053$; $dy/dx=(77.181-10(2.0053))/184.51=0.30962$.

Let required angle be $\alpha$: $\tan\alpha=0.30962\Rightarrow\alpha=17.2^\circ$ (1 d.p.).

Teacher notes

When the cannonball hits ground, $y=0$.

Time $=\text{distance}/\text{speed}$. The same flight time is the target’s travel time over distance $3000-d$.

Answer key

(i) $v^2\sin\theta\cos\theta/5$. (iii) $22.7^\circ$. (iv) $17.2^\circ$ (1 d.p.). No examination attribution is printed for this group.

Native diagrams