[tex]dane:\\v_{o} = 0\\h = 300 \ m\\g = 10\frac{m}{s^{2}}\\szukane:\\t = ?\\\\Rozwiazanie\\\\h = \frac{1}{2}gt^{2} \ \ /\cdot2\\\\gt^{2} = 2h \ \ /:g\\\\t^{2} = \frac{2h}{g}\\\\t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\cdot300 \ m}{10\frac{m}{s^{2}}}} = \sqrt{60 \ s^{2}}\\\\t \approx 7,7 \ s \ - \ czas \ spadania[/tex]
