Lösung 1.1:3

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The ball hits the ground when its height is zero, i.e. when
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<center> [[Image:1_1_3-1(2).gif]] </center>
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<math>h\left( t \right)=10-\frac{9.82}{2}t^{2}=0</math>
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<center> [[Image:1_1_3-2(2).gif]] </center>
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This second-degree equation has the solutions
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<math>t=\pm \sqrt{\frac{2\centerdot 10}{9.82}}</math>
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where the positive root is the time when the ball hits the ground.
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We obtain the ball's speed as a function of time as the time derivative of the height,
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<math>v\left( t \right)={h}'\left( t \right)=\frac{d}{dx}\left( 10-\frac{9.82}{2}t^{2} \right)=-9.82t</math>
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If we substitute the time when the ball hits the ground, we obtain the ball's speed at that instant,
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<math>\begin{align}
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& 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}} \\
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& =\sqrt{9.82\centerdot 2\centerdot 10}=-\sqrt{196.4}\approx -14.0 \\
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\end{align}</math>
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The minus sign shows that the speed is directed downwards, and the ball's speed is therefore
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<math>\text{14}.0\text{ }</math>
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m/s.

Version vom 11:58, 10. Okt. 2008

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


\displaystyle h\left( t \right)=10-\frac{9.82}{2}t^{2}=0

This second-degree equation has the solutions


\displaystyle t=\pm \sqrt{\frac{2\centerdot 10}{9.82}}


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\left( t \right)={h}'\left( t \right)=\frac{d}{dx}\left( 10-\frac{9.82}{2}t^{2} \right)=-9.82t


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


\displaystyle \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}


The minus sign shows that the speed is directed downwards, and the ball's speed is therefore \displaystyle \text{14}.0\text{ } m/s.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.1:3“