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Mechanics11/Seite10/Loesung3.17

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Angabe

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

Rechnung

h = 15m \frac{}{}

h = \frac{v_2}{2g}

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

v_0 = 17,16 \frac{m}{s}


t_s = \frac{v_0}{g}

t_s = \frac{17,6}{9,81}s

t_s = 1,75 s \frac{}{}


t_s = t_f \frac{}{}

t_s = 1,75 s \frac{}{}

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


Schülerlösung

geg: h = 15m

ges: ts; v0; tf


h = v2/2g

v0 = \sqrt[]{2g * h}

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

v0 = 17,1 m/s