Aus Online Mathematik Brückenkurs 2

Version vom 11:58, 10. Okt. 2008 (bearbeiten) Ian (Diskussion | Beiträge) ← Zum vorherigen Versionsunterschied		Version vom 12:58, 14. Okt. 2008 (bearbeiten) (rückgängig) Tek (Diskussion | Beiträge) K Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
	The ball hits the ground when its height is zero, i.e. when		The ball hits the ground when its height is zero, i.e. when
		+	{{Displayed math||<math>h(t) = 10-\frac{9\textrm{.}82}{2}t^{2} = 0\,\textrm{.}</math>}}
-	<math>h\left( t \right)=10-\frac{9.82}{2}t^{2}=0</math>	+	This quadratic equation has the solutions
-		+
-	This second-degree equation has the solutions	+
-		+
-		+
-	<math>t=\pm \sqrt{\frac{2\centerdot 10}{9.82}}</math>	+
		+	{{Displayed math||<math>t=\pm\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\,,</math>}}
	where the positive root is the time when the ball hits the ground.		where the positive root is the time when the ball hits the ground.
Zeile 14:		Zeile 11:
	We obtain the ball's speed as a function of time as the time derivative of the height,		We obtain the ball's speed as a function of time as the time derivative of the height,
-		+	{{Displayed math||<math>v(t) = h'(t) = \frac{d}{dt}\,\Bigl(10-\frac{9\textrm{.}82}{2}t^2\Bigr) = -9\textrm{.}82t\,\textrm{.}</math>}}
-	<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,		If we substitute the time when the ball hits the ground, we obtain the ball's speed at that instant,
		+	{{Displayed math||<math>\begin{align}
		+	v\Bigl(\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\,\Bigr)
		+	&= -9\textrm{.}82\cdot\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\\[5pt]
		+	&= -\sqrt{9\textrm{.}82^2\cdot\frac{2\cdot 10}{9\textrm{.}82}}\\[5pt]
		+	&= \sqrt{9\textrm{.}82\cdot 2\cdot 10}\\[5pt]
		+	&= -\sqrt{196\textrm{.}4}\\[5pt]
		+	&\approx -14\textrm{.}0\,\textrm{.}
		+	\end{align}</math>}}
-	<math>\begin{align}	+	The minus sign indicates that the speed is directed downwards, and the ball's speed is therefore 14.0 m/s.
-	& 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 12:58, 14. Okt. 2008

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

\displaystyle h(t) = 10-\frac{9\textrm{.}82}{2}t^{2} = 0\,\textrm{.}

This quadratic equation has the solutions

\displaystyle t=\pm\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\,,

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,

\displaystyle v(t) = h'(t) = \frac{d}{dt}\,\Bigl(10-\frac{9\textrm{.}82}{2}t^2\Bigr) = -9\textrm{.}82t\,\textrm{.}

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

\displaystyle \begin{align} v\Bigl(\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\,\Bigr) &= -9\textrm{.}82\cdot\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\\[5pt] &= -\sqrt{9\textrm{.}82^2\cdot\frac{2\cdot 10}{9\textrm{.}82}}\\[5pt] &= \sqrt{9\textrm{.}82\cdot 2\cdot 10}\\[5pt] &= -\sqrt{196\textrm{.}4}\\[5pt] &\approx -14\textrm{.}0\,\textrm{.} \end{align}

The minus sign indicates that the speed is directed downwards, and the ball's speed is therefore 14.0 m/s.