[tex]Dane:\\m = 300 \ g = 0,3 \ kg\\v_{o} = 0,1\frac{m}{s}\\g = 10\frac{m}{s^{2}}\\H = ?\\\\Rozwiazanie\\\\Z \ prawa \ zachowania \ energii\\\\E_{p} = E_{k}\\\\mgH = \frac{mv_{o}^{2}}{2} \ \ \ |:mg\\\\H = \frac{v_{o}^{2}}{2g}\\\\H = \frac{(0,1\frac{m}{s})^{2}}{2\cdot10\frac{m}{s^{2}}}=\frac{0,01\frac{m^{2}}{s^{2}}}{20\frac{m}{s^{2}}}\\\\\boxed{H = 0,0005 \ m}[/tex]
