Updated spacing between @ latex expressions
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| <title>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.</title> | |||
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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 | |||
| $$ v_x = v \cdot \cos{\varphi}\quad(1) $$ | |||
| And the vertical component depends as a function of time | |||
| $$ v_y(t) = v \cdot \sin{\varphi} - gt\quad(2) $$ | |||
| 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: | |||
| $$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2}\quad(3) $$ | |||
| $$ x(t) = vt \cdot \cos{\varphi}\quad(4) $$ | |||
| @@ -66,7 +66,7 @@ | |||
| c) From $(4)$ we express $t$: | |||
| $$ t = \frac{x}{v \cdot \cos{\varphi}} \quad(5) $$ | |||
| And we substitute in $(3)$ | |||
| − | $$ y(x) = x \tan | ||
| + | $$ y(x) = x \tan\varphi-\frac{gx^2}{2v^2\cos^2 \varphi}\quad(6) $$ | ||
| d) The maximum height is achieved when the velocity is directed horizontally $(v_y = 0)$: | |||
| $$ t_0 = \frac{v\sin\varphi}{g}\quad(7) $$ | |||
| Substituting $(7)$ into $(3)$: | |||
| $$ H = \frac{v^2}{2g}\operatorname{sin}^2\varphi\quad(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}\quad(9) $$ | |||
| Substituting $(9)$ into $(4)$: | |||
| $$ L=\frac{v^2}{g}\operatorname{sin}2\varphi\quad(10) $$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| @@ -82,7 +82,7 @@ | |||
| <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$<br> | |||
| − | c) $y = x \tan | ||
| + | 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)$<br> | ||
| 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$<br> | |||
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| Solutions of Savchenko Problems in Physics <br> | Solutions of Savchenko Problems in Physics <br> | ||
| <i><b>knowledge must be free</b></i> | <i><b>knowledge must be free</b></i> | ||
| </p> | </p> | ||
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| <h3> Statement </h3> | <h3> Statement </h3> | ||
| <p> | <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. | $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> | </p> | ||
| <h3>Solution</h3> | <h3>Solution</h3> | ||
| <p> | <p> | ||
| a) Since the external force (gravitational force) acts only in the vertical direction, the horizontal component of the velocity $v_x$ remains unchanged | 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}\quad(1) $$ | $$ v_x = v \cdot \cos{\varphi}\quad(1) $$ | ||
| And the vertical component depends as a function of time | And the vertical component depends as a function of time | ||
| $$ v_y(t) = v \cdot \sin{\varphi} - gt\quad(2) $$ | $$ v_y(t) = v \cdot \sin{\varphi} - gt\quad(2) $$ | ||
| b) Based on the obtained expressions for the speed $(1)$ and $(2)$, we obtain the dependence of the horizontal and vertical coordinates on time, respectively: | b) Based on the obtained expressions for the speed $(1)$ and $(2)$, we obtain the dependence of the horizontal and vertical coordinates on time, respectively: | ||
| $$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2}\quad(3) $$ | $$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2}\quad(3) $$ | ||
| $$ x(t) = vt \cdot \cos{\varphi}\quad(4) $$ | $$ x(t) = vt \cdot \cos{\varphi}\quad(4) $$ | ||
| @@ -66,7 +66,7 @@ | |||
| c) From $(4)$ we express $t$: | c) From $(4)$ we express $t$: | ||
| $$ t = \frac{x}{v \cdot \cos{\varphi}} \quad(5) $$ | $$ t = \frac{x}{v \cdot \cos{\varphi}} \quad(5) $$ | ||
| And we substitute in $(3)$ | And we substitute in $(3)$ | ||
| $$ y(x) = x \tan |
$$ y(x) = x \tan\varphi-\frac{gx^2}{2v^2\cos^2 \varphi}\quad(6) $$ | ||
| d) The maximum height is achieved when the velocity is directed horizontally $(v_y = 0)$: | d) The maximum height is achieved when the velocity is directed horizontally $(v_y = 0)$: | ||
| $$ t_0 = \frac{v\sin\varphi}{g}\quad(7) $$ | $$ t_0 = \frac{v\sin\varphi}{g}\quad(7) $$ | ||
| Substituting $(7)$ into $(3)$: | Substituting $(7)$ into $(3)$: | ||
| $$ H = \frac{v^2}{2g}\operatorname{sin}^2\varphi\quad(8) $$ | $$ H = \frac{v^2}{2g}\operatorname{sin}^2\varphi\quad(8) $$ | ||
| From the symmetry of the parabola, the time after which the body will again be on the ground$(y = 0)$, we find as: | From the symmetry of the parabola, the time after which the body will again be on the ground$(y = 0)$, we find as: | ||
| $$ T = 2t_0 = \frac{2v\sin\varphi}{g}\quad(9) $$ | $$ T = 2t_0 = \frac{2v\sin\varphi}{g}\quad(9) $$ | ||
| Substituting $(9)$ into $(4)$: | Substituting $(9)$ into $(4)$: | ||
| $$ L=\frac{v^2}{g}\operatorname{sin}2\varphi\quad(10) $$ | $$ L=\frac{v^2}{g}\operatorname{sin}2\varphi\quad(10) $$ | ||
| </p> | </p> | ||
| <h4>Answer</h4> | <h4>Answer</h4> | ||
| @@ -82,7 +82,7 @@ | |||
| <p> | <p> | ||
| a) $v_x= v\cos\varphi, v_y = v\sin\varphi-gt$<br> | a) $v_x= v\cos\varphi, v_y = v\sin\varphi-gt$<br> | ||
| b) $x = (v\cos\varphi)t, y = (v\sin{\varphi})t - gt^2/2$<br> | b) $x = (v\cos\varphi)t, y = (v\sin{\varphi})t - gt^2/2$<br> | ||
| c) $y = x \tan |
c) $y = x \tan\varphi-\frac{gx^2}{2v^2\cos^2 \varphi} =x\operatorname{tg}\varphi-\frac{gx^2}{2v^2}(\operatorname{tg}^2\varphi+1)$<br> | ||
| d) $T=\frac{2v}{g}\operatorname{sin}\varphi, H=\frac{v^2}{2g}\operatorname{sin}^2\varphi, L=\frac{v^2}{g}\operatorname{sin}2\varphi$<br> | d) $T=\frac{2v}{g}\operatorname{sin}\varphi, H=\frac{v^2}{2g}\operatorname{sin}^2\varphi, L=\frac{v^2}{g}\operatorname{sin}2\varphi$<br> | ||
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