Mechanics11/Seite10/Loesung3.17

Angabe

• geg.: Höhe (h = 15m)
• ges.: Anfangsgeschwindigkeit (v₀); Steigzeit (t_s); Fallzeit (t_f)

Rechnung

[math]h = 15m \frac{}{}[/math]

[math]h = \frac{v_2}{2g}[/math]

[math]v_0 = \sqrt[]{30 * 9.81}\frac{m}{s}[/math]

[math]v_0 = 17,16 \frac{m}{s}[/math]

[math]t_s = \frac{v_0}{g}[/math]

[math]t_s = \frac{17,6}{9,81}s[/math]

[math]t_s = 1,75 s \frac{}{}[/math]

[math]t_s = t_f \frac{}{}[/math]

[math]t_s = 1,75 s \frac{}{}[/math]

--FatF 12:25, 11 January 2008 (CET)

Schülerlösung

geg: h = 15m

ges: t_s; v₀; t_f

h = v₂/2g

v₀ = [math]\sqrt[]{2g * h}[/math]

v₀ = [math]\sqrt[] {2*9,81 m/s^2*15m}[/math]

v₀ = 17,1 m/s