Solution 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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{{Displayed math||<math>h(t) = 10-\frac{9\textrm{.}82}{2}t^{2} = 0\,\textrm{.}</math>}}
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<center> [[Image:1_1_3-2(2).gif]] </center>
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This quadratic equation has the solutions
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{{Displayed math||<math>t=\pm\sqrt{\frac{2\cdot 10}{9\textrm{.}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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{{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>}}
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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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{{Displayed math||<math>\begin{align}
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v\Bigl(\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\,\Bigr)
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&= -9\textrm{.}82\cdot\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\\[5pt]
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&= -\sqrt{9\textrm{.}82^2\cdot\frac{2\cdot 10}{9\textrm{.}82}}\\[5pt]
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&= \sqrt{9\textrm{.}82\cdot 2\cdot 10}\\[5pt]
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&= -\sqrt{196\textrm{.}4}\\[5pt]
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&\approx -14\textrm{.}0\,\textrm{.}
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\end{align}</math>}}
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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.

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