[tex]Dane:\\v_{o} = 0\\v = 10\frac{m}{s^{2}}\\g = 10\frac{m}{s^{2}}\\Szukane:\\h = ?\\\\Rozwiazanie\\\\Z \ zasady \ zachowania \ energii\\\\E_{p} = E_{k}\\\\mgh = \frac{mv^{2}}{2} \ \ /:mg\\\\h = \frac{v^{2}}{2g}\\\\h =\frac{(10\frac{m}{s})^{2}}{2\cdot10\frac{m}{s^{2}}}=\frac{100\frac{m^{2}}{s^{2}}}{20\frac{m}{s^{2}}}\\\\\boxed{h = 5 \ m}[/tex]
Odp. To ciało spada z wysokości h = 5 m.
