Mechanics11/Seite10/Loesung3.16

Lösung

a.

$v_0 = 20 \frac {m}{s}$

$h 0 = 1,5 m$ $a = - g$ Verzögerung am höchsten Punkt ist v=0 $v = v 0 + a * t$

$v-v_0 =a*t t= \frac {v-v_0}{a}$

$t=\frac {v_0}{g}=\frac {20\frac {m}{s}}{9,81\frac {m}{s^2}}=2,03s$

b

$E k = E p$

$0,5 * v 2 = g * h$

$h=\frac {v_0^2}{2g}$

$h=\frac {(20\frac {m}{s})^2}{2*9,81\frac {m}{s^2}}=20,4m$

c

ca: $-20\frac {m}{s}$

$v^2-v_0^2=2*a*x$

$a = g$

$x = h = (20,4 + 1,5) m = 21,9 m$

$v 0 = 0$

$v 2 = 2 * g * h$

$v=\sqrt []{2*g*h}=\sqrt []{2*9,81\frac {m}{s^2}*21,9m}=20,7\frac {m}{s}$ nach unten!!

$v = v 0 - g * t$

$v(1s)=10,2\frac {m}{s}$

$v(3s)=-9,43\frac {m}{s}$

$v(4s)=-19,2\frac {m}{s}$

--Lenny 13:09, 11 January 2008 (CET)