Answer
$4.52\space m/s,\space 59.4^{\circ}$
Work Step by Step
Please see the attached image first.
Let's apply equation 3.6b $V^{2}=u^{2}+2aS$ in the vertical direction to find the initial vertical velocity.
$\uparrow V^{2}=u^{2}+2aS$; Let's plug known values into this equation.
$(-8.9sin75^{\circ}\space m/s)^{2}=u_{y}^{2}+2(-9.8\space m/s^{2})(-3\space m)$
$u_{y}^{2}=74\space m^{2}/s^{2}-58.8\space m^{2}/s^{2}$
$u_{y}=3.9\space m/s$
The horizontal component of the velocity always remain constant. So we can write,
$u_{x}=8.9cos75^{\circ}m/s=2.3\space m/s$
By using the Pythagorean theorem we can get
$u=\sqrt {u_{x}^{2}+u_{y}^{2}}=\sqrt {(2.3\space m/s)^{2}+(3.89\space m/s)^{2}}=4.52\space m/s$
By using trigonometry, we can get
$tan\theta=\frac{3.89\space m/s}{2.3\space m/s}=>\theta=tan^{-1}(1.69)=59.4^{\circ}$
