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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>
$1.3.14.$ 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>
<h3>Solution</h3>
<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>
<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 class="exp">
$$v_{rel} = v_{y1} - v_{y2}$$
</p>
<p class="exp">
$$\fbox{$v_{rel} = -g \Delta t$}$$
</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:
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
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<meta name="description" content="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?">
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<meta property="og:description" content="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?">
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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>
$1.3.14.$ 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>
<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:
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
