Ile będzie spadać kamyk z odległości 1,5 m.

[latex]dane:\v_{o} = 0\h = 1,5 m\g = 10frac{m}{s^{2}}\szukane:\t = ?[/latex] [latex]s = frac{at^{2}}{2}\\s = h\a = g\\h = frac{gt^{2}}{2} |*2\\gt^{2}=2h /:g\\t^{2}=frac{2h}{g}\\t =sqrt{frac{2h}{g}} = sqrtfrac{2*1,5m}{10frac{m}{s^{2}}}=sqrt{0,3} s\\t approx0,5 s[/latex]