$1.3.14.$ 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 $\Delta t$?
Before viewing the solution, I recommend viewing the solution to problem 1.3.1.
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.
Subtracting each velocity component by component, we obtain that this relative velocity is (taking into account the equality $v_{x1} - v_{x2}$):
$$v_{rel} = v_{y1} - v_{y2}$$
$$\fbox{$v_{rel} = -g \Delta t$}$$
Using the formulas obtained in 1.3.6. By the time $\Delta t$, the first body will have relative coordinates:
$$x_{{rel}}=(v\cos\varphi)\Delta t$$
$$y_{{rel}}=(v\sin\varphi)\Delta t-g\Delta t^{2}/2$$
Let's take into account the motion of the second body relative to the first over time:
$$x_{{rel}}=(v\cos\varphi)\Delta t$$
$$y_{{rel}}=(v\sin\varphi)\Delta t-g\Delta t^{2}/2 - g \Delta t$$
$x_{{rel}}=(v\cos\varphi)\Delta t;$ $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 and equal in absolute value to $g \Delta t$