A ball is thrown from the top of a staircase which just touches the ceiling and finally hits the bottom of the steps. The initial speed of the ball is.
                     
 

         $4=\left(u\cos \alpha \right)t$                              …………… (1)
      $2=\frac{u^2{\sin }^2\alpha }{2g}$                                 …………… (2)
     $-3=\left(u\sin \alpha \right)t-\frac{1}{2}g{t}^2$              ……………. (3)
From (2)
       Using $\alpha =2\sqrt{g}$                             ……………….(4)
Substituting in (3), we get 
$$\begin{gathered} -3=6.3t-4.9t^2 \\ \Rightarrow 4.9t^2-6.3t-3=0 \\ \Rightarrow t=\frac{6.3\pm \sqrt{39.7-4\left(4.9\right)(-3)}}{2\left(4.9\right)} \\ \Rightarrow t=\frac{6.3\pm \sqrt{98.5}}{9.8} \\ \Rightarrow t=\frac{6.3+9.9}{9.8}=8.1s \end{gathered}$$
So, from (1),  $4=\left(u\cos \alpha \right)\left(8.1\right)$
$\Rightarrow u\cos \alpha =\frac{4}{8}\cong \frac{1}{2}\mathrm{...............}\left(5\right)$
From (4) and (5), we get 
$$\begin{gathered} u^2{\sin }^2\alpha +u^2{\cos }^2\alpha =40+\frac{1}{4} \\ \Rightarrow u=\sqrt{40.25}=6.3m{s}^{-1} \end{gathered}$$