Ile czasu będzie trwał spadek swobodny ciała z wysokości h = 1,8m?"

s=at*2/2 2s=at*2 t=√2s/a t=√1,8x2/10 t=0,6s

h(t)=[latex]h_{0} - frac{gt^{2} }{2} [/latex] 0=[latex]h_{0} - frac{gt^{2} }{2} [/latex] [latex]h_{0} = frac{g t^{2} }{2} [/latex] [latex]2 h_{0} =gt^{2}[/latex] [latex] frac{2 h_{0} }{g} = t^{2}[/latex] [latex] sqrt{frac{2 h_{0} }{g}} = t[/latex] [latex] sqrt{ frac{2*1,8}{10} }= sqrt {frac{3,6}{10} } = sqrt{0,36}=0,6 [/latex] Czas spadania wyniesie 0,6 sekundy.