[tex]dane:\\v_{o} = 0\\h = 20 \ m\\g = 10\frac{m}{s^{2}}\\szukane:\\t = ?\\\\Rozwiazanie\\\\Dla \ \ v_{o} = 0\\\\s = \frac{at^{2}}{2}\\\\h = s, \ \ g = a\\\\h = \frac{gt^{2}}{2} \ \ /\cdot2\\\\gt^{2} = 2h \ \ /:g\\\\t^{2} = \frac{2h}{g}\\\\t = \sqrt{\frac{2h}{g}}\\\\t = \sqrt{\frac{2\cdot20 \ m}{10\frac{m}{s^{2}}} } = \sqrt{4 \ s^{2}}= 2 \ s[/tex]
Odp. Kamień będzie spadał przez 2 sekundy.
