A projectile is projected with initial velocity $(6\hat{\mathrm{i}}+8\hat{\mathrm{j}})\mathrm{m}/\mathrm{s}.\text{ If }\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$, then what is the horizontal range of the projectile?

Initial velocity is $(6\hat{\mathrm{i}}+8\hat{\mathrm{j}})\mathrm{m}/\mathrm{s} (\mathrm{given}).$
Magnitude of velocity of projection is
$\mathrm{u}=\sqrt{{\mathrm{u}}_{\mathrm{x}}^2+{\mathrm{u}}_{\mathrm{y}}^2}=\sqrt{6^2+8^2}=10\mathrm{m}/\mathrm{s}$
Angle of projection
$\begin{array}{l}\tan \mathrm{\theta }=\frac{{\mathrm{u}}_{\mathrm{y}}}{{\mathrm{u}}_{\mathrm{x}}}=\frac{8}{6}\\ \Rightarrow \sin \mathrm{\theta }=\frac{4}{5}\text{ and }\cos \mathrm{\theta }=\frac{3}{5}\end{array}$
Now the horizontal range is,
$\begin{array}{l}\mathrm{R}=\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}}=\frac{{\mathrm{u}}^{2}2\sin \mathrm{\theta cos}\mathrm{\theta }}{\mathrm{g}}\\ =\frac{(10{)}^2\times 2\times (4/5)\times (3/5)}{10}=9.6\mathrm{m}\end{array}$