The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.

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Solutions of Savchenko Problems in Physics
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    <h3> Statement </h3>
    <p>
        $1.3.6.$ The gun is fired at an angle $\varphi$ to the horizon. Initial velocity of the projectile $v.$ The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the $x$ and $y$ coordinates on time; <br>c) the trajectory equation, i.e. the dependence of $y$ on $x$; <br>d) the flight time, the highest altitude and range of the projectile.
    </p>

    <h3>Solution</h3>
    <p>
        a) Since the external force (gravitational force) acts only in the vertical direction, the horizontal component of the velocity $v_x$ remains unchanged

$$v_x = v \cdot \cos{\varphi}\;(1)$$
And the vertical component depends as a function of time
$$v_y(t) = v \cdot \sin{\varphi} - gt\;(2)$$
b) Based on the obtained expressions for the speed $(1)$ and $(2)$, we obtain the dependence of the horizontal and vertical coordinates on time, respectively:
$$y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2}\;(3)$$

$$x(t) = vt \cdot \cos{\varphi}\;(4)$$
c) From $(4)$ we express $t$:
$$t = \frac{x}{v \cdot \cos{\varphi}} \;(5)$$
And we substitute in $(3)$
$$y(x) = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi}\;(6)$$
d) The maximum height is achieved when the velocity is directed horizontally $(v_y = 0)$:
$$t_0 = \frac{v\sin\varphi}{g}\;(7)$$
Substituting $(7)$ into $(3)$:
$$H = \frac{v^2}{2g}\operatorname{sin}^2\varphi\;(8)$$
From the symmetry of the parabola, the time after which the body will again be on the ground$(y = 0)$, we find as:
$$T = 2t_0 = \frac{2v\sin\varphi}{g}\;(9)$$
Substituting $(9)$ into $(4)$:
$$L=\frac{v^2}{g}\operatorname{sin}2\varphi\;(10)$$

    </p>

    <h4>Answer</h4>
    <p>
        a) $v_x= v\cos\varphi, v_y = v\sin\varphi-gt$<br>

b) $x = (v\cos\varphi)t, y = (v\sin{\varphi})t - gt^2/2$

c) $y = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi} =x\operatorname{tg}\varphi-\frac{gx^2}{2v^2}(\operatorname{tg}^2\varphi+1)$

d) $T=\frac{2v}{g}\operatorname{sin}\varphi, H=\frac{v^2}{2g}\operatorname{sin}^2\varphi, L=\frac{v^2}{g}\operatorname{sin}2\varphi$

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