[tex]dane:\\v_{o} = 30\frac{m}{s}\\g = 10\frac{m}{s^{2}}\\szukane:\\H = ?\\\\Rozwiazanie\\\\Z \ zasady \ zachowania \ energii\\\\E_{p} = E_{k}\\\\mgH = \frac{mv_{o}^{2}}{2} \ \ /:mg\\\\H = \frac{v_{o}^{2}}{2g}\\\\H = \frac{(30\frac{m}{s})^{2}}{2\cdot10\frac{m}{s^{2}}} = \frac{900\frac{m^{2}}{s^{2}}}{20\frac{m}{s^{2}}}\\\\H = 45 \ m[/tex]
Odp. Na wysokość 45 m.
