Answer
$v_x=11.78m/s$
$v_{y}=-12.9m/s$
$\theta=48^{\circ}$ below the horizontal
Work Step by Step
We know that
$v_x=vcos\theta$
$v_x=13m/s \times cos(25^{\circ})$
$v_x=11.78m/s$
and $v_{\circ}=v_{\circ y}-gt$
We plug in the known values to obtain:
$v_{y}=13m/s \times sin(25^{\circ})-(9.81m/s^2)(1.88s)$
$v_{y}=-12.9m/s$
Now $\theta^{\circ}=tan^{-1}(\frac{v_y}{v_x})$
We plug in the known values to obtain:
$\theta=tan^{-1}(\frac{-12.9m/s}{11.78})$
$\theta=48^{\circ}$ below the horizontal
