Question

A rifle that shoots bullets at${}^{\mathbf{460}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}}$is to be aimed at a target${}^{\mathbf{45}\mathbf{.}\mathbf{7}}$ 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?

Short answer

The rife barrel be pointed${}^{0.0484m}$ above the ground

Step-by-step solution

Given information

$$\begin{gathered} g=9.8m/s^2 \\ {v}_0=460m/s \\ R=45.7m \end{gathered}$$

Determining the concept of projectile motion 

This problem deals with the projectile motion of an object. In projectile motion,the motion experienced by an object or particle that moves under the action of gravity only.

In this case, the distance between the target and the rifle is given and they are levelled with each other, so using the formula for range of the projectile, the angle with which the projectile is fired can be found and using this, how high should the rifle barrel must be, can be calculated.

Formula:

The range of the particle in projectile motion is given by

$R=\frac{{\theta }_0^2\sin 2}{g}$ (i)

Determining the height of the rifle to be kept above the ground

Using equation (i),

$$\begin{gathered} R\times g=v_0^2\sin 2\theta \\ \sin 2\theta =\frac{R\times g}{v_0^2} \\ 2\theta ={\sin }^{-1}\left(\frac{R\times g}{v_0^2}\right) \\ \theta =\frac{{\sin }^{-1}\left({\displaystyle \frac{R\times g}{v_0^2}}\right)}{2} \\ \theta =\frac{{\sin }^{-1}\left({\displaystyle \frac{45.7\times 9.8}{{460}^2}}\right)}{2} \\ \theta =0.0606° \end{gathered}$$

Assume${}^{\mathrm{h}=\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{rifle}\mathrm{to}\mathrm{be}\mathrm{kept}\mathrm{above}\mathrm{the}\mathrm{ground}}$

$$\begin{gathered} \tan \theta =\left(\frac{h}{R}\right) \\ \tan 0.0606=\left(\frac{h}{45.7}\right) \\ 1.05\times {10}^{-3}=\frac{h}{45.7} \\ h=\left(1.05\times {10}^{-3}\right)\times 45.7 \end{gathered}$$

Thus

The rife barrel be pointed${}^{0.0484m}$