Aus Online Mathematik Brückenkurs 2

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.