# A rifle that shoots bullets at is to be aimed at 460m/s a target 45.7 m away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

Dec 12, 2015

$4.6 \text{cm}$

#### Explanation:

The situation looks like this: The mission here is to find $h$. We can do this by using the range $d$ to find $\theta$ then use some simple trigonometry to get $h$.

The range is given by:

$d = \frac{{v}^{2} \sin 2 \theta}{g}$

• assuming level ground

$\therefore \sin 2 \theta = \frac{\mathrm{dg}}{v} ^ 2$

$= \frac{45.7 \times 9.8}{460} ^ 2$

$\therefore \sin 2 \theta = 0.002116$

$\therefore 2 \theta = {0.1212}^{\circ}$

$\therefore \theta = {0.0606}^{\circ}$

Now looking at the diagram we can say that:

$\tan \theta = \frac{h}{45}$

$\therefore h = 45.7 \times \tan \theta$

$h = 45.5 \times 0.001$

$h = 0.0457 \text{m}$

$h = 4.57 \text{cm}$

So the rifle should be pointed at $4.6 \text{cm}$ above the target.

I have been asked to give the derivation for the range so here it is for anyone interested:

The key to these trajectory type questions is to treat the horizontal and vertical components of the motion separately and use the fact that they share the same time of flight.

The horizontal component of velocity is constant so we can write:

$v \cos \theta = \frac{d}{t}$

$\therefore t = \frac{d}{v \cos \theta} \text{ } \textcolor{red}{\left(1\right)}$

For the vertical component we can use the equation of motion

$v = u + a t$

This becomes:

$0 = v \sin \theta - \text{g"t_(max)" } \textcolor{red}{\left(2\right)}$

${t}_{\max}$ is the time taken to reach the maximum height. Since the path is symmetrical we can say:

${t}_{\max} = \frac{t}{2}$

Substitute this into $\textcolor{red}{\left(2\right)} \Rightarrow$

$0 = v \sin \theta - \frac{\text{g} t}{2}$

$\therefore t = \frac{2 v \sin \theta}{g} \text{ } \textcolor{red}{\left(3\right)}$

Now we can put $\textcolor{red}{\left(1\right)}$ equal to $\textcolor{red}{\left(3\right)}$ to eliminate $t \Rightarrow$

$\frac{d}{v \cos \theta} = \frac{2 v \sin \theta}{g}$

$\therefore d = \frac{{v}^{2} 2 \sin \theta \cos \theta}{g}$

There is a standard trig identity which states:

$\sin 2 \theta = 2 \sin \theta \cos \theta$

So substituting this into the expression for $\mathrm{dr} A r r$

$d = \frac{{v}^{2} \sin 2 \theta}{g}$

So there you go.