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| <meta name="author" content="Aliaksandr Melnichenka"> | | <meta name="author" content="Aliaksandr Melnichenka"> |
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | | <meta name="date" content="2023-10" scheme="YYYY-MM"> |
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| <title>From the same place with a time interval of \Delta t, two bodies are thrown with the same initial velocity v at an angle of \varphi to the horizon. How does the first body move relative to the second? Why does the relative velocity depend only on \Delta t?</title> | | <title>From the same place with a time interval of \Delta t, two bodies are thrown with the same initial velocity v at an angle of \varphi to the horizon. How does the first body move relative to the second? Why does the relative velocity depend only on \Delta t?</title> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $1.3.14.$ From the same place with a time interval of $\Delta t$, two bodies are thrown with the same initial velocity $v$ at an angle of $\varphi$ to the horizon. How does the first body move relative to the second? Why does the relative velocity depend only on $\Delta t$? | | $1.3.14.$ From the same place with a time interval of $\Delta t$, two bodies are thrown with the same initial velocity $v$ at an angle of $\varphi$ to the horizon. How does the first body move relative to the second? Why does the relative velocity depend only on $\Delta t$? |
| </p> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| <p>Before viewing the solution, I recommend viewing the solution to problem <a href="../1.3.1" target="_blank">1.3.1</a>.</p> | | <p>Before viewing the solution, I recommend viewing the solution to problem <a href="../1.3.1" target="_blank">1.3.1</a>.</p> |
| <p>Over equal intervals of time, the vertical component of both velocities will decrease by an amount equal to $\Delta v_y = g \Delta t$, while the horizontal component $v_x$ remains unchanged. Therefore, their relative velocity will be constant and equal to the vector difference of the initial velocities. | | <p>Over equal intervals of time, the vertical component of both velocities will decrease by an amount equal to $\Delta v_y = g \Delta t$, while the horizontal component $v_x$ remains unchanged. Therefore, their relative velocity will be constant and equal to the vector difference of the initial velocities. |
| </p> | | </p> |
| <p>Subtracting each velocity component by component, we obtain that this relative velocity is (taking into account the equality $v_{x1} - v_{x2}$): | | <p>Subtracting each velocity component by component, we obtain that this relative velocity is (taking into account the equality $v_{x1} - v_{x2}$): |
| </p> | | </p> |
| <p class="exp"> | | <p class="exp"> |
| $$v_{rel} = v_{y1} - v_{y2}$$ | | $$v_{rel} = v_{y1} - v_{y2}$$ |
| </p> | | </p> |
| <p class="exp"> | | <p class="exp"> |
| $$\fbox{$v_{rel} = -g \Delta t$}$$ | | $$\fbox{$v_{rel} = -g \Delta t$}$$ |
| </p> | | </p> |
| <p>Using the formulas obtained in <a href="../1.3.6" target="_blank">1.3.6</a>. By the time $\Delta t$, the first body will have relative coordinates: | | <p>Using the formulas obtained in <a href="../1.3.6" target="_blank">1.3.6</a>. By the time $\Delta t$, the first body will have relative coordinates: |
| </p> | | </p> |
| <p class="exp"> | | <p class="exp"> |
| $$x_{{rel}}=(v\cos\varphi)\Delta t$$ | | $$x_{{rel}}=(v\cos\varphi)\Delta t$$ |
| </p> | | </p> |
| <p class="exp"> | | <p class="exp"> |
| $$y_{{rel}}=(v\sin\varphi)\Delta t-g\Delta t^{2}/2$$ | | $$y_{{rel}}=(v\sin\varphi)\Delta t-g\Delta t^{2}/2$$ |
| </p> | | </p> |
| <p>Let's take into account the motion of the second body relative to the first over time: | | <p>Let's take into account the motion of the second body relative to the first over time: |
| </p> | | </p> |
| <p class="exp"> | | <p class="exp"> |
| $$x_{{rel}}=(v\cos\varphi)\Delta t$$ | | $$x_{{rel}}=(v\cos\varphi)\Delta t$$ |
| </p> | | </p> |
| <p class="exp"> | | <p class="exp"> |
| $$y_{{rel}}=(v\sin\varphi)\Delta t-g\Delta t^{2}/2 - g \Delta t$$ | | $$y_{{rel}}=(v\sin\varphi)\Delta t-g\Delta t^{2}/2 - g \Delta t$$ |
| </p> | | </p> |
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| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| <p>$x_{{rel}}=(v\cos\varphi)\Delta t;$ $y_{{rel}}=(v\sin\varphi)\Delta t$$-g\Delta t^{2}/2-g\Delta t\cdot t$, where $t$ is the time that has passed | | <p>$x_{{rel}}=(v\cos\varphi)\Delta t;$ $y_{{rel}}=(v\sin\varphi)\Delta t$$-g\Delta t^{2}/2-g\Delta t\cdot t$, where $t$ is the time that has passed |
| since the second body was ejected. The relative velocity is constant, directed vertically downwards | | since the second body was ejected. The relative velocity is constant, directed vertically downwards |
| and equal in absolute value to $g \Delta t$ | | and equal in absolute value to $g \Delta t$ |
| </p> | | </p> |
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