Mechanics11/Seite10/Loesung3.16

Lösung
a.

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

$$h_0 = 1,5m$$ $$a=-g $$  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: $$-20\frac {m}{s}$$

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

$$a=g$$

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

$$v_0=0$$

$$v^2=2*g*h$$

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