Aus Online Mathematik Brückenkurs 2

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.