[tex]Dane:\\m = 70 \ kg\\E_{p} = 14 \ kJ = 14\cdot1000 \ J = 14 \ 000 \ J\\g = 10\frac{N}{kg}\\Szukane:\\h = ?\\\\Rozwiazanie\\\\E_{p} = mgh \ \ \ |:mg\\\\h = \frac{E_{p}}{mg}\\\\h = \frac{14000 \ J}{70 \ kg\cdot10\frac{N}{kg}} = \frac{14000 \ N\cdot m}{700 \ N}\\\\\boxed{ h= 20 \ m}[/tex]
Odp. Na wysokości 20 metrów.
