In the laboratory rest frame:

Body velocity before collision: $\vec{v}_0 = (v_{0x}, v_{0y})$
Wall velocity: $\vec{u} = (-u, 0), \quad u > 0$


The relative velocity of the body in the reference frame of the wall:
\[
\vec{v}_0' = \vec{v}_0 - \vec{u}
\]

Componentwise:
\[
v_{0x}' = v_{0x} - (-u) = v_{0x} + u
\]
\[
v_{0y}' = v_{0y} - 0 = v_{0y}
\]

In the reference frame of the wall:
The wall is fixed (vertical)
The impact is absolutely elastic


Properties of elastic reflection from a vertical wall:
The normal component (horizontal) changes sign
The tangent component (vertical) is preserved

Before impact in the wall system: $\vec{v}_0' = (v_{0x} + u,\; v_{0y})$

After impact in the wall system:
\[
v_x'^{\text{после}} = -(v_{0x} + u)
\]
\[
v_y'^{\text{после}} = v_{0y}
\]

To go back to the laboratory system, you need to add the wall velocity to the velocity in the wall system.$\vec{u} = (-u, 0)$:

\[
\vec{v}_{\text{после}}^{\text{lab}} = \vec{v}' + \vec{u}
\]

Componentwise:
\[
v_x^{\text{after}} = (-v_{0x} - u) + (-u) = -v_{0x} - 2u
\]
\[
v_y^{\text{after}} = v_{0y} + 0 = v_{0y}
\]

The axis \(Oy\) is directed vertically upwards, the acceleration due to gravity \(g > 0\), the projections of the acceleration:
\[
a_x = 0, \quad a_y = -g.
\]

Motion is uniform in \(x\), uniformly accelerated in \(y\):
\[
x(t) = v_{0x} t
\]
\[
y(t) = v_{0y} t - \frac{g t^2}{2}
\]

From the equation of vertical motion \(y(t) = h\):
\[
v_{0y} t - \frac{g t^2}{2} = h
\]
Multiply by 2 and regroup:
\[
g t^2 - 2 v_{0y} t + 2h = 0
\]

Solution of a quadratic equation:
\[
t_{1,2} = \frac{2v_{0y} \pm \sqrt{(2v_{0y})^2 - 4g \cdot 2h}}{2g} = \frac{2v_{0y} \pm \sqrt{4v_{0y}^2 - 8gh}}{2g}
\]

Let the body be at the same height \(H\) at times \(t_1\) and \(t_2\). Then \(t_1\) and \(t_2\) are the roots of the equation \(y(t) = H\):
\[
v_{0y} t - \frac{g t^2}{2} = H \quad\Longrightarrow\quad \frac{g}{2} t^2 - v_{0y} t + H = 0
\]

Here are the coefficients:
\[
A = \frac{g}{2}, \quad B = -v_{0y}, \quad C = H.
\]

According to Vieta's theorem:
\begin{align*}
t_1 + t_2 &= -\frac{B}{A} = \frac{v_{0y}}{g/2} = \frac{2v_{0y}}{g} \\[2mm]
t_1 t_2 &= \frac{C}{A} = \frac{H}{g/2} = \frac{2H}{g}
\end{align*}

We express \(v_{0y}\) и \(H\):

\[
v_{0y} = \frac{g(t_1 + t_2)}{2}, \quad H = \frac{g t_1 t_2}{2}.
\]

The body returns to the ground \(y=0\) at the moment \(T > 0\). Equation:
\[
v_{0y} t - \frac{g t^2}{2} = 0 \quad\Longrightarrow\quad t\left(v_{0y} - \frac{g t}{2}\right) = 0
\]
Roots: \(t = 0\) (начало) и \(t = \frac{2v_{0y}}{g}\) (end of flight). Therefore:
\[
T = \frac{2v_{0y}}{g}
\]
We substitute \(v_{0y}\) из B3:
\[
T = \frac{2}{g} \cdot \frac{g(t_1 + t_2)}{2} = t_1 + t_2
\]

\[
T = t_1 + t_2
\]

The maximum height is reached at the moment\(t_{\text{vertex}} = \frac{v_{0y}}{g}\). We substitute \(v_{0y}\) из B3:
\[
t_{\text{верш}} = \frac{t_1 + t_2}{2}
\]

Height at this moment:
\[
y_{\max} = v_{0y} t_{\text{vertex}} - \frac{g}{2} t_{\text{vertex}}^2
\]
Подставляем \(v_{0y}\) и \(t_{\text{vertex}}\):
\begin{align*}
y_{\max} &= \frac{g(t_1 + t_2)}{2} \cdot \frac{t_1 + t_2}{2} - \frac{g}{2} \cdot \frac{(t_1 + t_2)^2}{4} \\
&= \frac{g(t_1 + t_2)^2}{4} - \frac{g(t_1 + t_2)^2}{8} \\
&= \frac{g(t_1 + t_2)^2}{8}
\end{align*}

Alternative method:
\[
y_{\max} = \frac{v_{0y}^2}{2g} = \frac{\left[\frac{g(t_1 + t_2)}{2}\right]^2}{2g} = \frac{g^2 (t_1 + t_2)^2}{4} \cdot \frac{1}{2g} = \frac{g(t_1 + t_2)^2}{8}.
\]

The \(x\) axis is directed from the starting point of the body to the point of fall without a wall.
Wall velocity: \(\vec{u}_{\text{wall}} = -u \,\vec{e}_x\) (\(u>0\)).
Before the impact the body has a velocity: \(v_x = v_{0x}\).

Relative to the wall, the velocity of the body before impact:
\[
v_{\text{rel}}^{\text{before}} = v_{0x} - (-u) = v_{0x} + u.
\]

After a perfectly elastic impact relative to the wall:
\[
v_{\text{rel}}^{\text{after}} = -(v_{0x} + u).
\]

In the laboratory system:
\[
v'_x = v_{\text{rel}}^{\text{after}} + (-u) = -(v_{0x} + u) - u = -v_{0x} - 2u.
\]

Let \(T = t_1 + t_2\) be the total flight time without a wall (from B4)
The wall at time \(t = 0\) is at the point of impact without a wall:
\[
X_{\text{ст}}(0) = v_{0x} T.
\]
Wall motion: \(X_{\text{ст}}(t) = v_{0x} T - u t\).
Body movement before impact: \(x(t) = v_{0x} t\).

Collision condition at the moment \(\tau\):
\[
v_{0x} \tau = v_{0x} T - u \tau.
\]
\[
u \tau = v_{0x} (T - \tau).
\]
\[
u = v_{0x} \cdot \frac{T - \tau}{\tau}.
\]

From (B7) for two experiments (\(t_1 < t_2\) — impact moments):
\[
u_i = v_{0x} \cdot \frac{T - t_i}{t_i}, \quad i = 1, 2.
\]

After the impact, the horizontal velocity of the body (from B6):
\[
v'_x(i) = -v_{0x} - 2u_i = -v_{0x} \left( 1 + 2\frac{T - t_i}{t_i} \right) = -v_{0x} \cdot \frac{2T - t_i}{t_i}.
\]

Impact coordinate: \(x_{\text{удар}}(i) = v_{0x} t_i\).
Remaining flight time: \(T - t_i\).

Movement after impact:
\[
\Delta x_i = v'_x(i) \cdot (T - t_i) = -v_{0x} \frac{2T - t_i}{t_i} (T - t_i).
\]

Final fall coordinate:
\[
X_i = x_{\text{удар}}(i) + \Delta x_i = v_{0x} \left[ t_i - \frac{(2T - t_i)(T - t_i)}{t_i} \right].
\]

Let's substitute \(T = t_1 + t_2\), \(T - t_1 = t_2\), \(T - t_2 = t_1\):

\[
X_1 = v_{0x} \left[ t_1 - \frac{(t_1 + 2t_2) t_2}{t_1} \right] = v_{0x} \cdot \frac{t_1^2 - t_1 t_2 - 2t_2^2}{t_1},
\]
\[
X_2 = v_{0x} \left[ t_2 - \frac{(t_2 + 2t_1) t_1}{t_2} \right] = v_{0x} \cdot \frac{t_2^2 - t_1 t_2 - 2t_1^2}{t_2}.
\]

Difference:
\[
X_1 - X_2 = v_{0x} \left[ \frac{t_1^2 - t_1 t_2 - 2t_2^2}{t_1} - \frac{t_2^2 - t_1 t_2 - 2t_1^2}{t_2} \right].
\]

Bringing to a common denominator and simplifying the numerator:
\[
X_1 - X_2 = v_{0x} \cdot \frac{2(t_1 + t_2)^2 (t_1 - t_2)}{t_1 t_2}.
\]

Since \(t_1 < t_2\), то \(t_1 - t_2 < 0\) и \(X_1 - X_2 < 0\).
Distance between drop points:
\[
L = |X_1 - X_2| = -(X_1 - X_2) = v_{0x} \cdot \frac{2(t_1 + t_2)^2 (t_2 - t_1)}{t_1 t_2}.
\]

Formula for the horizontal projection of the initial velocity:
\[
v_{0x} = \frac{L\, t_1 t_2}{2 (t_1 + t_2)^2 (t_2 - t_1)}.
\]
We substitute the numbers:
\[
v_{0x} = \frac{16 \cdot 2 \cdot 3}{2 \cdot (2 + 3)^2 \cdot (3 - 2)}
= \frac{96}{2 \cdot 25 \cdot 1}
= \frac{96}{50} = 1.92\ \text{m/s}.
\]


Vertical projection of initial velocity:
\[
v_{0y} = \frac{g (t_1 + t_2)}{2} = \frac{10 \cdot 5}{2} = 25\ \text{m/s}.
\]


\[
v_0 = \sqrt{v_{0x}^2 + v_{0y}^2}
= \sqrt{(1.92)^2 + 25^2}
= \sqrt{3.6864 + 625}
= \sqrt{628.6864} \approx 25.07\ \text{m/s}.
\]

The initial distance is given by
\[
S=v_{0x}(t_1+t_2)=1.92\text{ m/s}\cdot 5\text{ s}=9.6\text{ m}.
\]

For a vertical throw ($\alpha = 90^\circ$), the initial vertical velocity $v_{0y}=v_0$.
From the equation $v_y^2 = v_{0y}^2 - 2g y$ with $v_y=0$ we obtain
\[
0 = v_0^2 - 2g h_{\max} \quad\Longrightarrow\quad h_{\max}= \frac{v_0^2}{2g}.
\]

At zero initial altitude, the maximum range is achieved at $\alpha=45^\circ$:
\[
L_{\max}= \frac{v_0^2 \sin 90^\circ}{g}= \frac{v_0^2}{g}.
\]

Trajectory equation:
\[
y = x\tan\alpha - \frac{gx^2}{2v_0^2\cos^2\alpha}.
\]
We maximize $y$ for fixed $x$ with respect to $\alpha$:
\[
\frac{dy}{d\alpha}=0 \;\Longrightarrow\; \frac{x}{\cos^2\alpha}
- \frac{gx^2}{v_0^2}\cdot\frac{\sin\alpha}{\cos^3\alpha}=0.
\]
From here
\[
\tan\alpha = \frac{v_0^2}{gx}.
\]
We substitute into the trajectory equation:
\[
y = \frac{v_0^2}{g} - \frac{gx^2}{2v_0^2}.
\]
We bring it to the form $y=ax^2+bx+c$:
\[
a=-\frac{g}{2v_0^2},\qquad b=0,\qquad c=\frac{v_0^2}{2g}.
\]

The flight time $t$ is found from the equation
\[
0 = H + v_0\sin\alpha\,t - \frac{gt^2}{2},
\]
which has a positive root
\[
t = \frac{v_0\sin\alpha + \sqrt{v_0^2\sin^2\alpha+2gH}}{g}.
\]
Flight range
\[
R(\alpha)=v_0\cos\alpha\,t
= \frac{v_0\cos\alpha\bigl(v_0\sin\alpha+\sqrt{v_0^2\sin^2\alpha+2gH}\bigr)}{g}.
\]
The maximum $R(\alpha)$ for $\alpha$ is achieved when
\[
\tan\alpha^* = \frac{v_0}{\sqrt{v_0^2+2gH}}.
\]
Substitution gives
\[
R_{\max}= \frac{v_0\sqrt{v_0^2+2gH}}{g}.
\]

For elastic bounces off vertical walls, we use the mirror image method.
Horizontal displacement in the expanded image during flight $T$:
\[
\tilde{x}(T)=v_0\cos\alpha\,T
= \frac{v_0\cos\alpha\bigl(v_0\sin\alpha+\sqrt{v_0^2\sin^2\alpha+2gH}\bigr)}{g}.
\]
Maximum value $\tilde{x}(T)$ по $\alpha$coincides with $R_{\max}$ из C4:
\[
\tilde{x}_{\max}= \frac{v_0\sqrt{v_0^2+2gH}}{g}.
\]
Bounces occur when in the expanded image $\tilde{x}=\bigl(k+\frac12\bigr)l$ ($k=0,1,2,\dots$).
Maximum number of bounces:
\[
N_{\max}= \Bigl\lfloor \frac{\tilde{x}_{\max}}{l}+\frac12\Bigr\rfloor
=\Bigl\lfloor \frac{v_0\sqrt{v_0^2+2gH}}{gl}+\frac12\Bigr\rfloor.
\]

\[
\frac{v_0 \sqrt{v_0^2 + 2gH}}{g l}
= \frac{5 \cdot \sqrt{5^2 + 2\cdot 10\cdot 4}}{10 \cdot 2}
= \frac{5 \cdot \sqrt{25 + 80}}{20}
= \frac{5 \cdot \sqrt{105}}{20}.
\]


\[
\sqrt{105} \approx 10.24695, \quad
\frac{5 \cdot 10.24695}{20} = \frac{51.23475}{20} \approx 2.56174.
\]


\[
N_{\max} = \left\lfloor 2.56174 + \frac12 \right\rfloor
=\left\lfloor 3.06174 \right\rfloor
= 3.
\]