Answer to Question #94633 in Mechanics | Relativity for Ali Hamza

Question

A gun standing on sloping ground (see the figure below) fires up the slope. Show that the slant
range of the gun (measured along the slope) is
l =(2u*2cos*2θ/g cos α)(tan θ − tan α).
Here u is the magnitude of the initial velocity of the bullet

Answer 1

If $\theta$ and $\alpha$ are the angles between the horizontal and the velocity vector and the slope correspondingly, the horizontal range is:

$$R=L\space\text{cos}\alpha=v_xt=u\space\text{cos}\theta\space t,$$

where $t$ - time between the shot and the landing:

$$t=t_{up}+t_{down},$$

$$L\space\text{cos}\alpha=u\space\text{cos}\theta\space (t_{up}+t_{down})\space\space\space\space\space\space\space\space\space\space\space\space\space\space(1).$$

Time required for the bullet to reach the maximum height $H=v_y^2/2g$ above the horizontal is

$$t_{up}=\frac{v_y}{g}=\frac{u\space\text{sin}\theta}{g}.$$

Now equation (1) becomes

$$L\space\text{cos}\alpha=u\space\text{cos}\theta\space \Big(\frac{u\space\text{sin}\theta}{g}+t_{down}\Big).$$

Express $t_{down}$:

$$t_{down}=\frac{L\space\text{cos}\alpha}{u\space\text{cos}\theta}-\frac{u\space\text{sin}\theta}{g}\space\space\space\space\space\space\space(2).$$

Then the bullet falls down from height $H$ to $h$ - height of the point of landing above the horizontal:

$$H-h=\frac{gt_{down}^2}{2},\\ \space\\ \frac{u^2\text{sin}^2\theta}{2g}-h=\frac{gt_{down}^2}{2},\\ \space\\ h=\frac{u^2\text{sin}^2\theta}{2g}-\frac{gt_{down}^2}{2}.\space\space\space\space\space\space\space\space\space(3)$$

On the other hand, since height $h$ is nothing like

$$h=L\space\text{sin}\alpha,$$

we can substitute $t_{down}$ from (2) to (3) and get

$$L\space\text{sin}\alpha=\frac{u^2\text{sin}^2\theta}{2g}-\frac{g\Big(\frac{L\space\text{cos}\alpha}{u\space\text{cos}\theta}-\frac{u\space\text{sin}\theta}{g}\Big)^2}{2}.$$

Express $L$:

$$L=\frac{2u^2}{g}\Big[\frac{\text{cos}^2\theta\space\text{tan}\theta}{\text{cos}\alpha}-\frac{\text{sin}\alpha\space\text{cos}^2\theta}{\text{cos}^2\alpha}\Big]=\\ \space\\ =\frac{2u^2\text{cos}^2\theta}{g\space\text{cos}\alpha}[\text{tan}\theta-\space\text{tan}\alpha].$$