The horizontal and vertical displacement x and y of a projectile at a given time t are given by x = 6t metre and $y=8t-5t^2$  metre. The range of the projectile in metre is:

$x=\left(u\cos \theta \right)t=6t$

$y=\left(u\sin \theta \right)t-\frac{1}{2}g{t}^2=8t-5t^2$

Therefore,        $u\sin \theta =8$

                        $u\cos \theta =6$

Range $R=\frac{u^2\sin 2\theta }{8}=\frac{u^2\times 2\sin \theta \cos \theta }{g}$

$=\frac{2\left(u\sin \theta \right)\left(u\cos \theta \right)}{g}$

$=\frac{2\left(8\right)\left(6\right)}{10}=9.6m$