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Statement

$1.3.4.$ A free-falling body flew past point $A$ with velocity $v_A$. How fast will it fly past point $B$, which is $h$ below $A$?

Solution

From the Law of Conservation of Energy: $$ \frac{mv_A^2}{2} + mgh =\frac{mv_B^2}{2} $$ Where, $$ \fbox{$v_B = \sqrt{v_A^2 + 2gh}$} $$

Answer

$$v_B = \sqrt{v_A^2 + 2gh}$$