Answer
$2.5 m$ at an angle of $53^{\circ}$ with respect to x-axis.
Work Step by Step
$v_{x}=0.6m/s$
$v_{y}=0.8m/s$
Resultant $v=\sqrt (0.6^{2}+0.8^{2})=1 m/s$
In t=2.5 s, displacement=2.5 m.
$\theta=tan^{-1}(\frac{0.8}{0.6})=53^{\circ}$ with respect to x-axis.
