Solution 1.1:3

From Förberedande kurs i matematik 2

(Difference between revisions)
Jump to: navigation, search
(Ny sida: {{NAVCONTENT_START}} <center> Bild:1_1_3-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:1_1_3-2(2).gif </center> {{NAVCONTENT_STOP}})
Current revision (12:58, 14 October 2008) (edit) (undo)
m
 
(3 intermediate revisions not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
The ball hits the ground when its height is zero, i.e. when
-
<center> [[Bild:1_1_3-1(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
{{Displayed math||<math>h(t) = 10-\frac{9\textrm{.}82}{2}t^{2} = 0\,\textrm{.}</math>}}
-
{{NAVCONTENT_START}}
+
 
-
<center> [[Bild:1_1_3-2(2).gif]] </center>
+
This quadratic equation has the solutions
-
{{NAVCONTENT_STOP}}
+
 
 +
{{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.
 +
 
 +
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>}}
 +
 
 +
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>}}
 +
 
 +
The minus sign indicates that the speed is directed downwards, and the ball's speed is therefore 14.0 m/s.

Current revision

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.

Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_1.1:3"