Answer
$1130\space m/s,\space 37.7^{\circ}$
Work Step by Step
Let's apply equations 3.3a,b $V=u+at$ in the horizontal, and vertical directions to find the initial x and y- components of velocity.
$\rightarrow V=u+at$
$3775\space m/s=V_{0x}+5.1\space m/s^{2}\times565\space s$
$V_{0x}=893.5\space m/s$
$\uparrow V=u+at$
$4816\space m/s=V_{0y}+7.3\space m/s^{2}\times565\space s$
$V_{0x}=691.5\space m/s$
By using the Pythagorean theorem, we can get
Initial Velocity $V_{0}=\sqrt {(893.5\space m/s)^{2}+(691.5\space m/s)^{2}}=1130\space m/s$
By using trigonometry, we can get
$tan\theta=\frac{V_{0y}}{V_{0x}}=\frac{691.5\space m/s}{893.5\space m/s}=>\theta=tan^{-1}(0.774)=37.7^{\circ}$
