Answer
45.5 m/s
Work Step by Step
Let's apply equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in vertical direction to the projectile.
$\uparrow S=ut+\frac{1}{2}at^{2}$ Let's plug known values into this equation.
$0=Vsin30^{\circ}t+\frac{1}{2}\times (-9.8\space m/s^{2})t^{2}$
$t=\frac{V}{9.8}s-(1)$
Let's apply the equation $S=ut$ in the horizontal direction to the projectile.
$\rightarrow S=ut$ ; Let's plug known values into this equation.
$183\space m=Vcos30^{\circ}t-(2)$
(1)=>(2),
$183\space =\sqrt 3\frac{V}{2}\times\frac{V}{9.8}\space $
$V=\sqrt {\frac{183\times9.8\times2}{\sqrt 3}}m/s=45.5\space m/s$
Initial velocity of the projectile = 45.5 m/s