A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is v, the angle of inclination of the mountain is \alpha, and the angle of fire relative to the horizon is \beta?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#1.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.3.8.$ A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is $v$, the angle of inclination of the mountain is $\alpha$, and the angle of fire relative to the horizon is $\beta$?

    <h3>Solution</h3>
    <p>
        The equations of motion of the projectile can be written as follows:

$${x}={v}_{0}{t}\cos{\beta}$$

$${y}={v}_{0}{t}\sin{\beta}-\frac{{g}{t}^{2}}{2}$$
Substitute into the equations the coordinates of the target $x = L; \;y = L \tan \alpha$
$${L=v_{0}t\cos\beta}$$

$$Ltg\alpha=v_{0}t\sin\beta-\frac{gt^{2}}{2}$$
Let us express time from the first equation of the last system of equations and substitute its value into the second equation
$${t=\frac{L}{v_{0}\cos\beta}}$$

$${Ltg\alpha=v_{0}\frac{L}{v_{0}\cos\beta}sin\beta-\frac{g}{2}\frac{L^{2}}{v_{0}^{2}\cos^{2}\beta} }$$
Where
$${v}_{0}=\sqrt{\frac{{gL}\cos\alpha}{2\cos\beta\sin(\beta-\alpha)}}$$
We express $L$,
$$L = \frac{ 2\cos\beta\sin(\beta-\alpha)\cdot v^2_0}{g\cos\alpha}$$
And we find the flight range along the wall:
$$l = \frac{L}{\cos \alpha}$$

$$\fbox{$l = \frac{ 2v^2_0}{g} \frac{ \cos\beta\sin(\beta-\alpha)}{\cos^2\alpha}$}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$L=\frac{2v^2}g\frac{\cos^2\beta}{\cos\alpha}(\text{tg}\beta-\text{tg}\alpha)$$
    </p>


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