Translated 1.4.1-1.4.18

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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#1.4">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $1.4.11.$ The body is dropped over the slab at a height $h$ from it. The slab moves vertically upwards at a speed of $u$. Determine the time between two consecutive impacts of the body on the slab. The blows are absolutely elastic.
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <p>Let's move to the reference frame associated with the plate. At the initial moment of time, the relative velocity is $u$ and is directed downwards.</p>
+<p>From the law of conservation of energy, we find the velocity of the body $v$ at the moment before the impact</p>
+<p class="exp">
+$$ \frac{mv^2}{2} = \frac{mu^2}{2} + mgh $$
+</p>
+<p>Where is the desired velocity </p>
+<p class="exp">$$v = \sqrt{u^2+2gh}$$</p>
+<p>In this inertial reference frame, after an elastic rebound, the relative velocity will only change direction.</p>
+<p>A body thrown vertically will land after time</p>
+<p class="exp">
+$$ t = \frac{2v}{g} $$
+</p>
+<p>Time sought:</p>
+<p class="exp">
+$$ \fbox{$t = \frac{2\sqrt{u^2 + 2gh}}{g}$} $$
+</p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$t = 2\sqrt{u^2/g^2 + 2h/g}$$
+ </p>
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