A ball is thrown upwards from the ground with an initial speed u. The ball is at a height of 80 m at two times, the time interval being 6s. Then u is $(Takeg=10m{s}^{-2})$

Here, $a=g=-10m{s}^{-2}ands=80m$

Substituting the values egh, in  $s=ut+\frac{1}{2}a{t}^2,$ we have

$80=ut-5t^2$

or $5t^2-ut+80=0$

or $t=\frac{u+\sqrt{u^2-1600}}{10}$

and $t=\frac{u-\sqrt{u^2-1600}}{10}$

Now, it is given that

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

or $\frac{\sqrt{u^2-1600}}{5}=6$

or $\sqrt{u^2-1600}=30$

or $u^2-1600=900$

$\therefore u^2=2500$

or $u=\pm 50m{s}^{-1}$

Ignoring the negative sign, we have

$u=50m{s}^{-1}$