A ball is thrown upwards from the ground with an initial speed of  $u$. At two instants of time, having an interval of 6s between the to instants, the ball is at a height of 80 m from the ground. Find  $u\text{\hspace{0.17em}\hspace{0.17em}in\hspace{0.17em}}m{s}^{-1}$.Take $g=10{\text{ms}}^{-2}$.

Consider the upward direction as positive, then  $u=+um{s}^{-1},a=g=-10ms{}^{-2}$and  $s=+80m$
Since,  $s=ut+\frac{1}{2}a{t}^2,$ so we get 
$80=ut-5t^2$       $s=80\text{m}$
$\Rightarrow 5t^2-ut+80=0$
 $\Rightarrow t=\frac{u+\sqrt{u^2-1600}}{10}$ and $\frac{u-\sqrt{u^2-1600}}{10}$     
Now it is given that these two instants, when the ball is at 80 m from the ground have an interval of 6s between.
So,

$\frac{u+\sqrt{u^2-1600}}{10}-\frac{u-\sqrt{u^2-1600}}{10}=6$


$\Rightarrow \frac{\sqrt{u^2-1600}}{5}=6$
$\Rightarrow \sqrt{u^2-1600}=30$
$\Rightarrow u^2-1600=900$

$\Rightarrow u^2=2500$

$\Rightarrow u=\pm 50m{s}^{-1}$

$\Rightarrow \left|u\right|=50m{s}^{-1}$