\begin{frame}
  \frametitle{Exercises}

%   \begin{exampleblock}{}
%     Assume that a ball is dropped,
%     and we have the following measurements:
%     \begin{itemize}
%       \item height at time $0s$ is $490$m
%       \item height at time $2s$ is $472$m
%       \item height at time $4s$ is $414$m
%     \end{itemize}
%     Find a quadratic function for the height of the ball after time $t$.\\
%     When does the ball hit the ground?
%   \end{exampleblock}

  Formula for the height:
  \begin{exampleblock}{}
  \begin{malign}
    h(t) = -5t^2 + t + 490
  \end{malign}
  \end{exampleblock}
  \pause\bigskip
  
  When does the ball hit the ground? \pause When the height is $0$:
  \pause
  \begin{talign}
    &-5t^2 + t + 490 = 0 \quad
    \mpause[1]{\implies t^2 - \frac{t}{5} - 98 = 0}
  \end{talign}
  \pause\pause
  Solving the quadratic formula:
  \begin{talign}
    t = \frac{1}{10} \pm \sqrt{(\frac{1}{10})^2 + 98}
      \mpause[1]{= \frac{1}{10} \pm \sqrt{\frac{1}{100} + \frac{9800}{100}}}
      \mpause[2]{= \frac{1}{10} \pm \frac{\sqrt{9801}}{10}}
  \end{talign}
  \pause\pause\pause
  We know $100^2 = 10000$ \pause and $(100 - n)^2 = 10000 - 200n + n^2$.\\
  \pause
  Thus $\sqrt{9801} = 99$.\pause
  \begin{talign}
    t = \frac{1}{10} \pm \frac{99}{10} 
    \mpause[1]{\quad\implies\quad t = 10 \text{\;\; or \;\;} t = - \frac{98}{10} }
  \end{talign}
  \pause\pause
  Thus the ball hits the ground after $10$ seconds.
\end{frame}