[tex]Dane:\\v_{o} = 0\\v = 20\frac{m}{s}\\g = 10\frac{m}{s^{2}}\\Szukane:\\h = ?[/tex]
Rozwiązanie
Korzystamy z prawa zachowania energii:
[tex]E_{p} = E_{k}\\\\mgh = \frac{mv^{2}}{2} \ \ \ /:m\\\\gh = \frac{v^{2}}{2} \ \ \ |\cdot2\\\\2gh = v^{2} \ \ \ /:2g\\\\\underline{h = \frac{v^{2}}{2g}}\\\\h = \frac{(20\frac{m}{s})^{2}}{2\cdot10\frac{m}{s^{2}}} = \frac{400\frac{m^{2}}{s^{2}}}{20\frac{m}{s^{2}}}\\\\\boxed{h = 20 \ m}[/tex]
Odp. Kulka spada z wysokości h = 20 m.
