Kinematics

$t_c =\dfrac{l}{v_1 + v_2}$; $x_c = \dfrac{lv_1}{v_1 + v_2}$.

The dependence of the coordinates of the bodies on time is given by the formulas: $x_1 = v_1 t; \quad x_2 = l - v_2 t.$ The meeting ($x_1 = x_2$) will occur after time $t = l/(v_1 + v_2)$ from the start of motion at a point located at a distance $lv_1/(v_1 + v_2)$ from point A. The graphs of the dependence of the coordinates of the bodies on time are shown below ($\tan \alpha = v_1$; $\tan \beta = -v_2 < 0$, since $v_2$ is directed in the opposite direction to the one taken as positive). The moment of meeting corresponds to point C of intersection of the graphs.

$t_c = \dfrac{l - v_2 t_0}{v_1 - v_2}$; $x_c = \dfrac{(l - v_2 t_0) v_1}{v_1 - v_2}$.

Use the graphs of the dependence of the coordinates of the bodies on time (the following figure).

$t_2 = 6$ h.

Let $u$ be the speed of the current, and $v$ be the speed of the boat relative to the water, we can write: $t_1 = \dfrac{s}{v+u}; \tag{1}$ $t = \dfrac{s}{u}. \tag{2}$ To determine the time $t_2 = s/(v-u)$ taken by the boat for the return trip, it is convenient to rewrite equations (1) and (2) as follows: $\dfrac{1}{t_1} = \dfrac{v}{s} + \dfrac{u}{s}; \tag{3}$ $\dfrac{1}{t} = \dfrac{u}{s}. \tag{4}$ Subtracting twice equation (4) from equation (3), we get $\dfrac{1}{t_1} - \dfrac{2}{t} = \dfrac{v}{s} - \dfrac{u}{s} = \dfrac{1}{t_2},$ whence $t_2 = \dfrac{t t_1}{t - 2t_1}.$

1.5 min.

$n = 100$.

If the person’s velocity were directed opposite to the motion of the escalator, then the faster he walked, the fewer steps he would count (but no fewer than $ n$. In our case, the directions of the person’s and the escalator’s motion coincide.

Let $v$ be the speed of the escalator; $l$ its length, and $n$ the number of steps on a stationary escalator. The number of steps per unit length of the escalator is $n / l$.

Therefore, if the person walks with speed $u$ relative to the escalator, the time spent on the escalator is $l / (v + u)$, and the distance he travels (relative to the escalator) is $ul / (v + u)$.

Thus, the number of steps the person counts is:
$n_1 = \dfrac{ul}{v + u} \cdot \dfrac{n}{l} = \dfrac{u}{v + u} n.$

Similarly, in the second case (if the walking speed is tripled), he counts:
$n_2 = \dfrac{3ul}{v + 3u} \cdot \dfrac{n}{l} = \dfrac{3u}{v + 3u} n.$
Hence, we obtain the following system of equations:

From this, by eliminating the ratio $v/u$, we find:
$n = \dfrac{2n_1 n_2}{3n_1 - n_2} = 100.$

$u = 7.5$ km/h; $v = 17.5$ km/h.

$u = 4$ km/h; $v = 16$ km/h.

Let's choose the raft (water) as the reference frame. In this reference frame, the boat moves down and up the river with the same speed. This means that the time for the boat to move away from the raft is equal to the time to approach it; thus, the boat also returned in 3/4 h. During the elapsed 1.5 h, the raft traveled a distance $s_1-s_2 = 6 \text{ km}.$ Consequently, the speed of the current (speed of the raft relative to the bank) is $u = 6 / 1.5 = 4~ \text{km/h}.$ The speed of the boat relative to the water $v = \dfrac{s_1}{t} - u = 16~\text{km/h}.$

This solution illustrates how important a successful choice of reference frame is in kinematics.

$t = 2$ min.

The speed of the cyclist in the reference frame associated with the column is $v_2 - v_1$ when moving towards the head of the column, and $v_2 + v_1$ when returning. Therefore $t = \dfrac{l}{v_2 - v_1} + \dfrac{l}{v_2 + v_1} = \dfrac{2lv_2}{v_2^2 - v_1^2}.$

600 m/s.

$v_2 = v_1 \tan \alpha \approx 14.4$ m/s.

For the raindrops not to leave a trace on the rear window, the direction of the raindrop velocity vector relative to the car must make an angle with the horizon no greater than $60^\circ$. The relative velocity is composed of the vertical component, equal to the falling speed of the drop $v_2$, and the horizontal component, equal to the speed of the car $-v_1$ and directed in the opposite direction (following figure). From the velocity triangle, it is clear that $v_2 = v_1 \tan \alpha$.

Identical.

$v = 174$ km/h; northwest at an angle $4^\circ 27'$ to the meridian.

Let $v_0 = s/t$ be the speed of the aircraft relative to the Earth. It is directed, according to the condition, along the meridian; then $\vec{v}_0 = \vec{v} + \vec{u}$. By the cosine theorem $v^2 = u^2 + v_0^2 - 2uv_0 \cos 150^\circ.$ The angle $\beta$ between $\vec{u}$ and $\vec{v}_0$ is found from the sine theorem: $\sin \beta / u = \sin 150^\circ / v.$

a) and b) - a straight line forming an angle $\alpha = \arctan(u/v)$ with the direction of motion of the board. In case b), the trace may not reach the edge of the board.

Consider the motion of the chalk in the coordinate system associated with the board. Since the friction force is directed along the velocity vector of the chalk relative to the board, it cannot change the direction of the chalk's motion and only reduces its speed.

$\beta = \arcsin(\dfrac{v}{u} \sin \alpha)$.

Point C (see figure below) is the meeting place of the ship and the torpedo. $AC = vt$, $CB = ut$, where $t$ is the travel time of the torpedo. According to the sine theorem $\dfrac{AC}{\sin \beta} = \dfrac{BC}{\sin \alpha}$ or $\dfrac{vt}{\sin \beta} = \dfrac{ut}{\sin \alpha}.$ Hence $\beta = \arcsin(\dfrac{v}{u} \sin \alpha).$

$u = v / \cos \alpha$.

For the same small time interval $\Delta t$, the slider moves by $AB = \Delta l$, and the cord is pulled by the length $AC = \Delta l \cos \alpha$ (see figure) ($\angle BCA$ can be considered right due to the smallness of $\Delta l$). Therefore, we can write $\Delta l / u = \Delta l \cos \alpha / v$, whence $u = v / \cos \alpha$, i.e., the speed of pulling the rope is equal to the projection of the slider's velocity onto the direction of the rope.

$u = v / \cos \alpha$.

The projection of the velocity $v$ of the load onto the direction of each cable must be equal to the speed $u$ of the cable (see problem 15).

$v_B = v_A \cot \alpha = 17.3$ cm/s; $OB = \sqrt{l^2 - v_A^2 t^2}$; $v_B = v_A^2 t / \sqrt{l^2 - v_A^2 t^2}$.

See solution to problem 15. At any moment in time, the projections of the velocities $v_A$ and $v_B$ of the ends of the rod (see figure below) onto the axis of the rod are equal to each other, otherwise the rod would have to shorten or lengthen. Thus, $v_A \cos \alpha = v_B \sin \alpha$, whence $v_B = v_A \cot \alpha$.

a) 40 m/s;

b) 0;

c) $20\sqrt{2}$ m/s.

The velocity of any point on the track relative to the Earth at any given moment is composed of the velocity of the tank relative to the Earth and the velocity of the track link relative to the tank.

1. $v_{\text{avg}} = \dfrac{2v_1 v_2}{v_1 + v_2} = 48$ km/h;

2. $v_{\text{avg}} = \dfrac{2v_1(v_2+v_3)}{2v_1 + v_2 + v_3} = 40$ km/h.

1. The average speed $v_{\text{avg}}$ is numerically equal to the ratio of the total path traveled by the body to the time of motion: $v_{\text{avg}} = s_{\text{total}}/t.$ Time $t = t_1 + t_2 = \dfrac{s/2}{v_1} + \dfrac{s/2}{v_2} = \dfrac{s(v_1+v_2)}{2v_1 v_2},$ therefore $v_{\text{avg}} = s/t = \dfrac{2v_1 v_2}{v_1 + v_2}.$

2.Hint. The car traveled the second half of the path with an average speed equal to $(v_2+v_3)/2$.

$v_1 = v_{\text{avg}} \dfrac{n+1}{2} = 54$ km/h; $v_2 = v_{\text{avg}} \dfrac{n+1}{2n} = 36$ km/h.

The speeds will be the same. The travel time of the second ball is less.

Sample graphs of the speed of the balls are shown in the following figure. Since the paths traveled by the balls are equal, as seen from the graph (on the graph, the path is numerically equal to the areas of the shaded figures), $t_2 < t_1$.

$\sqrt{v^2 - u^2}/v$.

If the wind blows along the line AB, the total flight time is $t_1 = \dfrac{l}{v+u} + \dfrac{l}{v-u} = \dfrac{2l v}{v^2 - u^2},$ and the average speed $v_{\text{avg}1} = \dfrac{2l}{t_1} = \dfrac{v^2 - u^2}{v}$ where $l$ is the distance between cities. If the wind is perpendicular to the line AB, the speed of the aircraft relative to the wind must be directed at an angle to the line AB, such that the drift is compensated (see figure). The flight speed in this case is constant and $v_{\text{avg}2} = \sqrt{v^2 - u^2}$ (the flight time in this case is $t_2 = 2l/\sqrt{v^2 - u^2}$). $v_{\text{avg}1}/v_{\text{avg}2} = (v^2 - u^2)/(v \sqrt{v^2 - u^2}) = \sqrt{v^2 - u^2}/v.$

$v_{\text{max}} \approx 16.2$ m/s. The graph of the train's speed is shown in the following figure.

2:1; see the following figure.

The path traveled by the train after the car was detached is equal to the area of the rectangle OABC; the path traveled by the car is the area of the triangle OAC.

$v = 7$ m/s; see the following figure.

From the graph, it is clear that the paths traveled by the passerby and the train are equal at the moment when $v = 2v_0$, where $v_0$ is the speed of the passerby.

Figure below.

The tangent of the angle of inclination of the tangent to the graph of the dependence of the coordinate (or path) of the body on time is numerically equal to the speed of the body. Therefore, the straight line representing the graph $x(t)$ between $t_1$ and $t_2$ at the points corresponding to moments $t_1$ and $t_2$ will be tangent to the parabolas describing the motion of the body before moment $t_1$ and after moment $t_2$, which are the graphs $x(t)$ or $s(t)$ before moment $t_1$ and after moment $t_2$ respectively. At moments $t=0$ and $t=t_3$, the graphs $s(t)$ and $x(t)$ have a horizontal tangent.

See the following figure.

The straight line between points $t_1$ and $t_2$ is tangent to the curves $x(t)$ and $s(t)$ describing the motion of the body before moment $t_1$ and after moment $t_2$.

$s = \dfrac{v_0^2}{a} + l$; $v_{\text{avg}} = \dfrac{v_0^2 + al}{v_0 + \sqrt{v_0^2 + 2al}}$.

Let's introduce a coordinate system with the origin at point A and the x-axis directed along the velocity of the body at point A (see figure). When the body reaches point B, its coordinate will be $-l$. The body reaches point B only if its acceleration $a$ is directed opposite to the velocity $v_0$. Therefore, the dependence of velocity $v$ on time is expressed by the formula $v = v_0 - at$. At the moment $t_1 = v_0/a$, the velocity becomes zero. At this moment, the body is at the maximum distance from point A. For $t > t_1$, $v < 0$. Substituting $t = t_1$ into the kinematic equation of motion: $x = v_0 t - at^2/2$, we find $x_{\text{max}} = v_0^2 / (2a)$. The path $s$ traveled by the body during the entire movement is $s = 2x_{\text{max}} + l = 2\dfrac{v_0^2}{2a} + l = \dfrac{v_0^2}{a} + l.$ To find the average speed $v_{\text{avg}}$, you need to know the time of motion of the body from point A to point B. It can be found by substituting the coordinate value of point B into the equation for x, i.e., $x = -l$: $t_2 = (v_0 + \sqrt{v_0^2 + 2al})/a$ (the second root is negative, so it must be discarded). Then the average speed of the body is $v_{\text{avg}} = s/t_2 = (v_0^2 + al)/(v_0 + \sqrt{v_0^2 + 2al}).$

See the following; $v_1 = \tan \alpha$. Point $t_2$ corresponds to the maximum coordinate of the body. Before moment $t=t_1$, the motion is uniform, for $t > t_1$ - uniformly accelerated with negative acceleration (i.e., acceleration directed opposite to velocity $v$).

$t_3 = t_2 + \sqrt{t_2(t_2-t_1)}$.

Equating the paths traveled by the bodies by the moment $t_3$ of meeting: $a_1 t_3^2 / 2 = a_2 (t_3 - t_1)^2 / 2. \tag{1}$ From the above figure, it is clear that $a_1/a_2 = (t_2-t_1)/t_2$ ($a_1 = \tan \alpha$; $a_2 = \tan \beta$). Substituting the ratio $a_1/a_2$ into equation (1) and solving the resulting equation for $t_3$, we find $t_3 = t_2 + \sqrt{t_2(t_2-t_1)}$. The second root is discarded, as the condition $t_3 > t_2$ must be met. The graphs of motion are shown in the following.

During the second second.

$v = \sqrt{La}$.

The time to transport the cargo will be minimum if the average speed of movement of the trolley is maximum, which is obvious. The latter, under the conditions of this problem (start and end of motion occur with constant acceleration), can only be in the case where the trolley moves the first half of the path with acceleration $+a$, and the second - with acceleration $-a$ (see figure). Thus, we can write the following relations: $v\tau/2 \cdot 2 = L/2$, $\tau/2 = v/a$, whence $v = \sqrt{La}$.

$a = 10$ m/s$^2$; $v = 300$ m/s.

$a = \dfrac{2l(t_1-t_2)}{t_1 t_2 (t_1+t_2)} = 3.20$ m/s$^2$; $v_0 = l/t_1 + at_1/2 \approx 13.6$ m/s.

The paths traveled by the train by moments $t_1$ and $t_1+t_2$ are respectively equal: $l = v_0 t_1 + at_1^2/2$ and $2l = v_0(t_1+t_2) + a(t_1+t_2)^2/2$.

$v = \dfrac{5l}{6t}$.

The problem is solved most simply if we use the reversibility of motion and choose the boundary between the two segments as the origin of coordinates. Then the ball has an initial velocity - the sought value $v$, while the acceleration in the direction towards the initial point is $a$, and towards the final point is $(-a)$. Writing the equation of motion for the paths traveled by the ball, $l = vt + at^2/2$; $l = v(3t) - a(3t)^2/2$, and eliminating $a$, we find the answer.

$\tau_n = 0.48$ s.

Let the entire length of the board be $5l$, then we can write that the time to pass the first segment is $\tau = \sqrt{2l/a}$. The times to pass the beginning and end of the n-th segment past the observer are respectively equal: $t_1 = \sqrt{\dfrac{2(n-1)l}{a}} = \tau \sqrt{n-1}$ and $t_2 = \sqrt{\dfrac{2nl}{a}} = \tau \sqrt{n},$ whence $\tau_n = t_2 - t_1 = \tau (\sqrt{n} - \sqrt{n-1}).$

$t = 1.8 \times 10^{-3}$ s; $a \approx 2.21 \times 10^5$ m/s$^2$; $v \approx 282$ m/s; $s = 32$ cm; $v_1 = 40$ m/s.

When determining at what depth the bullet's speed decreased by three times, use the formula $s = v_0^2 / (2a)$. Starting from the point where the speed decreased by three times, the remaining path will be $3^2 = 9$ times less than the total path.

$v_0 = 45$ cm/s; $a = 30$ cm/s$^2$.

1st method. The dependence of the coordinate of the body along the inclined plane on time is expressed by the formula $x = v_0 t - at^2/2$. Hence $t^2 - \dfrac{2v_0}{a} t + \dfrac{2x}{a} = 0.$ Since $t_1$ and $t_2$ are the roots of this equation for $x=l$, then, according to Vieta's theorem, $t_1 + t_2 = \dfrac{2v_0}{a}\quad\text{and}\quad t_1 t_2 = \dfrac{2l}{a}.$ From the obtained system of equations, it is easy to find $v_0$ and $a$.

2nd method. The dependence of the ball's speed on time is expressed by the formula $v = v_0 - at$. At time moments $t=t_1$ and $t=t_2$, the ball had speeds equal in magnitude and opposite in direction: $v_1 = -v_2$. But $v_1 = v_0 - at_1\quad\text{and}\quad v_2 = v_0 - at_2,$ therefore
$v_0 - at_1 = -v_0 + at_2,\quad\text{or}\quad 2v_0 = a(t_1+t_2).\tag{1}$
Since the ball moves with uniform acceleration, its average speed over time $t_1$ is $v_{\text{avg}1} = \dfrac{v_0+v_1}{2} = \dfrac{2v_0 - at_1}{2}.$ Therefore, $l = v_{avg1} t_1 = (2v_0 - at_1)\dfrac{t_1}{2},\quad\text{or}\quad 2v_0 - at_1 = \dfrac{2l}{t_1}.\tag{2}$
Solving equations (1) and (2) together, we find $a = \dfrac{2l}{t_1 t_2} (\dfrac{1}{t_1} + \dfrac{1}{t_2}).$

$t_1 \approx 0.45$ s; $t_2 \approx 0.023$ s; $s_1 \approx 4.9$ m; $s_2 \approx 40$ m.

$\tau = \sqrt{2l/g}(\sqrt{n_2} - \sqrt{n_1})$.

$h \approx 195$ m.

$t \approx 5.45$ s; $h \approx 145$ m.

Let $h$ be the height from which the body fell, and $t$ be the time of its fall from this height, we can write the following equations: $gt^2/2 = h$; $g(t-1)^2/2 = (2/3)h$. Solving them together, we get $t$ and $h$. The problem can be solved more shortly if we use the formula for the path traveled by the body in the t-th second: $s_t = \dfrac{g}{2}(2t-1);$ $\dfrac{g}{2}(2t-1) = \dfrac{1}{3} \dfrac{gt^2}{2}.$

$v_0 = \sqrt{2gh}$.

$\tau \approx 1$ s.

Let: $\tau$ be the time interval between the detachment of the 1st and 2nd drops, $t$ be the time from the moment of detachment of the 2nd drop, $d$ be the distance between the drops after time $t$ (from the moment of detachment of the 2nd drop) and considering that the drops move relative to each other uniformly, we can write the equation to determine $\tau$: $d = \dfrac{g\tau^2}{2} + g\tau t.$

$v_0 = \dfrac{g}{2} t_0 + \dfrac{8h}{t_0}$; $t = \sqrt{t_0^2 + \dfrac{8h}{g}}$.

See problem 38. Solve the problem using another method.

5 s; 75 m below point B.

The relative speed of the bodies is constant and equal to $2v_0$. Both bodies approach each other with this speed. Therefore, the time to meeting is $l/(2v_0) = 5$ s.

$h = (3/4)h_{\text{max}} = (3/4)v_0^2/(2g)$.

The problem can be solved beautifully if you realize that the moment of meeting divides the ascent time, as well as the fall time of the bodies, into equal parts. The paths traveled by a body falling without initial velocity in successive equal time intervals relate as 1:3. Therefore, the meeting will occur at 3/4 of the height.

$t = \dfrac{v_0}{g} - \dfrac{\tau}{2} = 1.75$ s; $h = \dfrac{v_0^2}{2g} - \dfrac{g\tau^2}{8} \approx 19.3$ m.

See problem 38. Solve the problem graphically as well.

$t \approx 3.4$ s.

Let's take the coordinate axis directed vertically upwards, with the origin on the Earth's surface. Then the kinematic equation of motion of the object has the form $y = y_0 + v_0 t - g\dfrac{t^2}{2},$ where $y_0 = a\tau^2/2$ and $v_0 = a\tau.$ Substituting the coordinate value of the object at the moment of fall $y=0$ into this equation, we find $t$.

$l = v_0^2 + 2uv_0 / (2g)$; $h_{\text{max}} = (u+v_0)^2 / (2g)$; $t = 2(v_0+u)/g$.

The kinematic equation of motion of the body is conveniently written in the coordinate system associated with the moving aerostat. Here $v_0+u$ is the initial velocity of the body; $g$ is the acceleration; $h_{\text{max}}$ is the coordinate of the body when the velocity of the body relative to the aerostat is zero; $t$ is its coordinate at the moment $t_1$, when its velocity $v$ relative to the Earth is zero. Since $v = v_0 - gt$, then $v=0$ at $t_1 = v_0/g$. At $t = \tau$, the coordinate of the body is zero.

$v_0 = 7$ m/s.

1st method. Taking the Earth's surface as the origin, let's write the equations of motion for both bodies:
$x_1 = H - \dfrac{gt^2}{2}; \tag{1}$
$x_2 = H - h + v_0 t - \dfrac{gt^2}{2}.\tag{2}$
At the moment of landing $x_1 = x_2 = 0$. Substituting the values $x_1$ and $x_2$ into equations (1) and (2) and solving them for $v_0$, we find $v_0 = h\sqrt{\dfrac{g}{2H}}.$

2nd method. Both bodies approach each other, moving uniformly relative to each other. The initial distance between them is $h$. Therefore, the velocity $v_0$ can be found from the equation of uniform motion $v_0 = h/t$, where $t$ is the fall time of the first body from height H. Since $t = \sqrt{2H/g}$. We find $v_0 = h/\sqrt{2H/g} = h\sqrt{g/(2H)}$.

$v_0 = \dfrac{H-h}{2h} \sqrt{2gh} \approx 7$ m/s.

$v_0 = \sqrt{\dfrac{Lg \cos^2 \alpha }{2 \sin \alpha}}$.

Writing the coordinate equations for the stone (see figure): $x = v_0 t;\quad y =\dfrac{gt^2}{2}$ and considering that at the moment of fall of the stone $x = L \cos \alpha,\quad y = L \sin \alpha,$ we get after eliminating $t$, that $v_0 = \sqrt{\dfrac{Lg \cos^2 \alpha}{2 \sin \alpha}}$.

4.9 m.

The ascent time of the ball is 1 s.

$\tan \alpha = v \sqrt{2/(gh)}$; $s = v \sqrt{2h/g}$.

$\tan \alpha = s/h$ (see figure).

The fall time of the first body, which did not experience a collision, is less than the fall time of the second body. The speeds of the bodies at the moments when they were at the same height are identical.

During an elastic collision, the magnitude of the velocity is conserved, only its direction changes.

0.64 s; 0.52 m.

In the coordinate system associated with the lift, the acceleration of the bolt is $9.8 \text{ m/s}^2 + 2 \text{ m/s}^2 = 11.8 \text{ m/s}^2$. Relative to the shaft, the initial velocity of the bolt is 2.4 m/s, and the acceleration is $9.8 \text{ m/s}^2$.

$\Delta h \approx 56.4$ m; the bodies move apart with constant velocity.

In the vertical direction, both bodies move away from each other with a velocity equal to $v_0 \cos \alpha - (-v_0 \cos \alpha) = 2v_0 \cos \alpha$.

If the initial velocities of the bodies are $\vec{v}_0$ and $\vec{u}_0$, then after time $t$ their velocities are respectively $\vec{v} = \vec{v}_0 + \vec{g}t$ and $\vec{u}=\vec{u}_0+\vec{g}t$ ($\vec{g}$ is the acceleration vector of the bodies). The velocity of the second body relative to the first is $\vec{v}_{\text{rel}} = \vec{u} - \vec{v} = \vec{u}_0 - \vec{v}_0.$

$\tan \alpha = \sqrt{3}$; $\alpha = 60^\circ$.

1st method. Both bodies can meet on the line AO (see figure above) at point C (it can also lie below the line BO).

Let's resolve the initial velocity $v_0$ of the body thrown from point B into horizontal $v_{0x}$ and vertical $v_{0y}$ components: $v_{0x} = v_0 \cos \alpha,\quad v_{0y} = v_0 \sin \alpha.$ From the start of motion until the moment of meeting, the elapsed time is: $t = l/v_{0x} = l/(v_0 \cos \alpha).\tag{1}$ During this time, body A will descend by $H-h = gt^2/2, \tag{2}$ and body B will rise to height $h = v_{0y} t - gt^2/2 = v_0 \sin \alpha t - gt^2/2. \tag{3}$ Solving equations (2) and (3) together, we find $H = t v_0 \sin \alpha. \tag{4}$ Substituting the value of $t$ from expression (1) into equation (4), we get $\tan \alpha = \dfrac{H}{l},$ i.e., the throwing angle $\alpha$ does not depend on the initial velocity, only the location of the meeting point C on the line AO depends on its value.

2nd method. Let's switch to the coordinate system associated with the body falling from point A. Here the velocity of the body thrown from point B is constant (see the previous problem). Therefore, it is clear that the bodies will meet if the velocity vector of the body is directed towards point A, i.e., $\tan \alpha = {l}. \tag{4}$

$v = \sqrt{v_0^2 - 2gh}$ and does not depend on the throwing angle.

$t = v (\sin \alpha - \cos \alpha \tan \beta)/g \approx 0.75$ s.

$h_1:h_2:h_3 = 3:2:1$; $l_1:l_2:l_3 = \sqrt{3}:\sqrt{2}:1$.

Maximum ascent height $h = v_0^2 \sin^2 \alpha / (2g)$ and range of water flight $l = t v_0 \cos \alpha$, where $t$ is the time of water movement, equal to twice the ascent time to height $h$; $t = 2v_0 \sin \alpha / g$, whence $l = v_0^2 \sin 2\alpha / g$.

$t = \sqrt{2d/g}$; does not depend on the angle.

Along the grooves, the weights will move under the action of the component of gravity $mg \cos \alpha$ and, consequently, with acceleration $a = g \cos \alpha$; the length of the groove $l = d/\cos \alpha$ (see the above figure). The travel time $t$ of the weight is $t = \sqrt{\dfrac{2l}{a}} = \sqrt{\dfrac{2d}{g}}.$ From the formula, it is clear that $t$ does not depend on the angle.

$H_{\text{max}} \approx 2.8$ m.

Obviously, the stone is thrown upwards at an angle, because if it were thrown vertically, its speed after 0.5 s would be $v_0 - gt = 5.1$ m/s. The height to which the stone rises, $H_{\text{max}} =\dfrac{v_{0y}^2}{2g}, \tag{1}$ where $v_{0y}$ is the vertical component of $v_0$; $v_{0y}$ can be determined from the following system of equations: $v_0^2 = v_x^2 + v_{0y}^2, \tag{2}$ $v^2 = v_x^2 + (v_{0y} - gt)^2, \tag{3}$ where $v_x$ is the horizontal projection of $v_0$. Subtracting equation (2) from equation (3), we get $v_{0y} = \dfrac{v_0^2 - v^2 + g^2 t^2}{2gt}.\tag{4}$ Substituting expression (4) into equation (1), we get $H_{\text{max}} = \dfrac{(v_0^2 - v^2 + g^2 t^2)^2}{8g^3 t^2}.$

A sphere with radius $v_0 t$, whose center lies below the initial point by $gt^2/2$.

As the reference body, choose a body that starts falling down from the same point at the moment the balls are thrown. Relative to it, the balls will move (disperse in possible directions) uniformly with speed $v_0$.

$v_0 = \sqrt{\dfrac{gL \cos \alpha}{2 \cos \beta \sin(\beta - \alpha)}}$; $\beta_0 = \dfrac{\pi}{4} + \dfrac{\alpha}{2}$.

Let's choose the coordinate $xy$ system, whose origin is at the location of the gun (see the above figure). Then the kinematic equations of motion of the projectile in coordinates x, y will be expressed as follows: $x = v_0 t \cos \beta,\quad y = v_0 t \sin \beta - gt^2/2.$ Substituting the coordinates of the target $x=L$; $y=L \tan \alpha$ into these equations, eliminating $t$, we get $v_0 = \sqrt{\dfrac{gL \cos \alpha}{2 \cos \beta \sin(\beta - \alpha)}}.$ The range $l$ of the projectile along the slope is $l = L/\cos \alpha$. Therefore, the formula we obtained can be written as: $v_0 = \sqrt{\dfrac{gl \cos^2 \alpha}{2 \cos \beta \sin(\beta - \alpha)}}.$ Hence $l = \dfrac{2 v_0^2 \cos \beta \sin(\beta - \alpha)}{g \cos^2 \alpha}.$ This expression is maximum at the maximum product $\cos \beta_0 \sin(\beta_0 - \alpha) = \dfrac{1}{2} [\sin(2\beta_0 - \alpha) - \sin \alpha].$ Therefore, $l$ is maximum at the maximum value $\sin(2\beta_0 - \alpha) = 1$ or $\beta_0 = \dfrac{\pi}{4} + \dfrac{\alpha}{2}$ (for $\alpha = 0$ we get the answer $\beta_0 = \pi/4 = 45^\circ$).

$t = 2\sqrt{2gh}/g$; does not depend on the angle.

Let's take the coordinate system xOy, whose origin is located at the point of the first contact of the body with the inclined plane, the x-axis is directed along the inclined plane, and the $y$-axis is perpendicular to it (see figure below). With this choice of coordinate system, the solution of the problem is significantly simplified. The initial velocity $v_0$ of the body after the collision will be $v_0 = \sqrt{2gh}$ and forms an angle $\alpha$ with the $y$-axis. The equation of motion of the body along the $y$-axis will be expressed as follows: $y = v_{0y} t + a_y t^2 / 2,$ where $v_{0y} = v_0 \cos \alpha\text{ and } a_y = -g \cos \alpha$ are the projections of the velocity $v_0$ and acceleration $g$ onto the $y$-axis. At the moment the body falls onto the inclined plane $y=0$. Substituting $y=0$ into the equation written above, we get $t = 2v_{0y}/a_y = 2v_0/g = 2\sqrt{2gh}/g.$ The fall time does not depend on the angle of inclination of the plane. This time will remain unchanged for subsequent falls after the first reflection onto the inclined plane.

$4H\sqrt{2}$; $s_1 = 8H \sin \alpha$; $s_1:s_2:s_3 = 1:2:3$.

See the previous problem. The distance from the place of the first impact to the second will be determined from the relation $x = v_{0x} t + a_x t^2 / 2$, where $v_{0x} = v_0 \sin \alpha$ and $a_x = g \sin \alpha$ (see figure). Substituting the value $t$ obtained in the previous problem, we find that $ \begin{aligned} s_1 &= v_0 \dfrac{2v_0}{g} \sin \alpha + \dfrac{g}{2} (\dfrac{2v_0}{g})^2 \sin \alpha \\ &= \dfrac{4v_0^2}{g} \sin \alpha \\ &= 8H \sin \alpha; \end{aligned} $ and for the components of the velocity of the body $v_{2x}$ and $v_{2y}$ at the moment of the second collision, respectively: $v_{2y} = v_{0y} - a_y t = v_0 \cos \alpha - g \dfrac{2v_0}{g} \cos \alpha = -v_{0y};$ $ \begin{aligned} v_{2x} &= v_{0x} + a_x t \\ &=v_0 \sin \alpha + g \dfrac{2v_0}{g} \sin \alpha \\ &= 3v_0 \sin \alpha, \end{aligned} $ and at the moment of the n-th collision: $v_{ny} = -v_{0y},$ $v_{nx} = (2n-1)v_0 \sin \alpha.$ The average velocities along the $x$-axis between impacts are respectively equal: $2v_0 \sin \alpha,\quad 4v_0 \sin \alpha,\quad 6v_0 \sin \alpha,$ i.e., they relate as 1:2:3, and since the time of motion between impacts is the same, then $s_1:s_2:s_3... = 1:2:3.$

$\tau \le 0.09$ s.

The error in determining the time between the shot and the arrival of the echo is $2\tau$ seconds. This leads to an error in determining the distance traveled by sound, equal to $2\tau c$ meters. The path traveled by sound is equal to twice the distance to the mountain $2x$, therefore the relative error in determining this distance is $2\tau c / (2x) = \tau c / x$. This is also the relative error in determining the distance to the mountain. Since $\tau c / x \le 0.03$, then $\tau \le 0.03 (x/c)$. Considering that $x \ge 1000$ m, we find that the error in determining the moments of the shot and the arrival of the echo should not exceed 0.09 s.

$\tau \approx 1.77$ s.

If we do not take into account the time of sound propagation, we get that the depth of the well is $h_1 = g\tau^2/2$. Taking into account the time of sound propagation $\tau = t_1 + t_2$, where $t_1 = \sqrt{2h/g}$ is the time of the stone's fall, and $t_2 = h/c$ is the time of sound propagation ($h$ is the true depth of the well): $\tau = \dfrac{h}{c} + \sqrt{\dfrac{2h}{g}}.\tag{1}$ The error in measuring the depth of the well is $\dfrac{\Delta h}{h} =\dfrac{h_1 - h}{h} = 0.05.$ Hence $h = 0.95 h_1 = 0.95 \dfrac{g\tau^2}{2}.$ Substituting this expression into equation (1) and solving the resulting equation, we find $\tau = \dfrac{2(1 - 1/\sqrt{0.95})c}{0.95 g} \approx 1.77\text{ s}.$