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Aus Online Mathematik Brückenkurs 2

Version vom 14:21, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 1.1:3 moved to Solution 1.1:3: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 11:58, 10. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	The ball hits the ground when its height is zero, i.e. when
-	<center> [[Image:1_1_3-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+	<math>h\left( t \right)=10-\frac{9.82}{2}t^{2}=0</math>
-	<center> [[Image:1_1_3-2(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	This second-degree equation has the solutions
		+
		+
		+	<math>t=\pm \sqrt{\frac{2\centerdot 10}{9.82}}</math>
		+
		+
		+	where the positive root is the time when the ball hits the ground.
		+
		+	We obtain the ball's speed as a function of time as the time derivative of the height,
		+
		+
		+	<math>v\left( t \right)={h}'\left( t \right)=\frac{d}{dx}\left( 10-\frac{9.82}{2}t^{2} \right)=-9.82t</math>
		+
		+
		+	If we substitute the time when the ball hits the ground, we obtain the ball's speed at that instant,
		+
		+
		+	<math>\begin{align}
		+	& v\left( \sqrt{\frac{2\centerdot 10}{9.82}} \right)=-9.82\centerdot \sqrt{\frac{2\centerdot 10}{9.82}}=-\sqrt{9.82^{2}\centerdot \frac{2\centerdot 10}{9.82}} \\
		+	& =\sqrt{9.82\centerdot 2\centerdot 10}=-\sqrt{196.4}\approx -14.0 \\
		+	\end{align}</math>
		+
		+
		+	The minus sign shows that the speed is directed downwards, and the ball's speed is therefore
		+	<math>\text{14}.0\text{ }</math>
		+	m/s.

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Version vom 11:58, 10. Okt. 2008

The ball hits the ground when its height is zero, i.e. when

ht=10−2982t2=0 

This second-degree equation has the solutions

t=982210 

where the positive root is the time when the ball hits the ground.

We obtain the ball's speed as a function of time as the time derivative of the height,

vt=ht=ddx10−2982t2=−982t 

If we substitute the time when the ball hits the ground, we obtain the ball's speed at that instant,

v982210=−982982210=−9822982210=982210=−1964−140 

The minus sign shows that the speed is directed downwards, and the ball's speed is therefore 140 m/s.