[tex]dane:\\v_{o} = 6\frac{m}{s}\\X = 2H\\g = 10\frac{m}{s^{2}}\\szukane:\\t = ?\\\\Rozwiazanie\\\\X = v_{o}\sqrt{\frac{2H}{g}}\\\\X = 2H\\\\2H = v_{o}\sqrt{\frac{2H}{g}} \ \ |()^{2}\\\\4H^{2} = v_{o}^{2}\cdot\frac{2H}{g} \ \ /:4H\\\\H = \frac{v_{o}^{2}}{2g} = \frac{(6\frac{m}{s})^{2}}{2\cdot10\frac{m}{s^{2}}}=1,8 \ m[/tex]
[tex]t_{s} = t = \sqrt{\frac{2H}{g}}\\\\t = \sqrt{\frac{2\cdot1,8 \ m}{10\frac{m}{s^{2}}}} = \sqrt{0,36 \ s^{2}}} = 0,6 \ s[/tex]
Odp. Czas lotu tego ciała t = 0,6 s.
