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+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.2.42.$ With what force does the cobra press on the ground when it is preparing to jump, rises vertically upwards at a constant speed $v$? The mass of the snake is $m$, its length is $l$.
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="2.2.42.jpg"
+ loading="lazy" width="230" />
+ <figcaption>
+ Forces acting on cobra
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+Let's consider small part of cobra with the length $\Delta x$ and mass $\Delta m$:
+\begin{equation*}
+\Delta P = \Delta m \cdot v
+\end{equation*}
+Since the snake is homogeneous
+\begin{equation*}
+\Delta m = \frac{m}{L} \cdot \Delta x
+\end{equation*}
+
+By newton's second law:
+\begin{equation*}
+F=\frac{\Delta P}{\Delta t} = \frac{\Delta m}{\Delta t} \cdot v
+\end{equation*}
+
+$$F = \frac{m}{L} \cdot \frac{\Delta x}{\Delta t} \cdot v = \frac{mv^2}{L}$$
+
+By definition of speed
+\begin{equation*}
+\frac{\Delta x}{\Delta t} = v
+\end{equation*}
+
+Whence the force associated with the acceleration of the centre of mass
+\begin{equation*}
+F = \frac{\Delta P}{\Delta t} = \frac{mv^2}{L}
+\end{equation*}
+
+Let's take everything together and find the support reaction force
+\begin{equation*}
+N = mg + \frac{mv^2}{L} = m\left(g + \frac{v^2}{L}\right)
+\end{equation*}
+
+\begin{equation*}
+\boxed{N = m\left(g + \frac{v^2}{L}\right)}
+\end{equation*}
+
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$N = m\left(g + \frac{v^2}{L}\right)$$
+ </p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Almaskhan Arsen<br>
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