[tex]Dane:\\v_{o} =0\\t = 2,4 \ s\\Szukane:\\h = ?\\v = ?[/tex]
Rozwiązanie
[tex]Dla \ \ g = 10\frac{m}{s^{2}}[/tex]
Głębokość wykopu h:
[tex]h = \frac{1}{2}gt^{2}\\\\h = \frac{1}{2}\cdot10\frac{m}{s^{2}}\cdot(2,4 \ s)^{2}\\\\\boxed{h = 28,80 \ m}[/tex]
Prędkość kamienia w momencie upadku v:
Dla v₀ = 0
[tex]v = g\cdot t\\\\v = 10\frac{m}{s^{2}}\cdot2,4 \ s\\\\\boxed{v = 24\frac{m}{s}}[/tex]
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[tex]Dla \ \ g = 9,81\frac{m}{s^{2}}\\\\h = \frac{1}{2}gt^{2}\\\\h = \frac{1}{2}\cdot9,81\frac{m}{s^{2}}\cdot(2,4 \ s)^{2}}\\\\\boxed{h = 28,25 \ m}[/tex]
[tex]v = g\cdot t\\\\v = 9,81\frac{m}{s^{2}}\cdot2,4 \ s\\\\\boxed{v = 23,54\frac{m}{s}}[/tex]
