Answer
$21.9\space m/s,\space 40^{\circ}$
Work Step by Step
Please see the attached image first.
Let's apply the equation $S=ut$ in the horizontal direction to find flight time.
$\rightarrow S=ut$ ; Let's plug known values into this equation.
$51\space m=23cos43^{\circ}m/s\times t$
$t=3.04\space s$
Let's apply the equation $V=u+at$ in the vertical direction to find the initial vertical velocity.
$\uparrow V=u+at$ ; Let's plug known values into this equation.
$-23sin43^{\circ}m/s=V_{0}+(-9.8)(3.04)\space m/s$
$V_{0}=14.1\space m/s$
By using the Pythagorean theorem, we can get.
$Initial Velocity = \sqrt {V_{0}^{2}+u_{0}^{2}}=\sqrt {(14.1\space m/s)^{2}+(23cos43^{\circ}m/s)^{2}}=21.9\space m/s$
By using trigonometry, we can get.
$tan\theta=\frac{14.1\space m/s}{23cos43^{\circ}m/s}=>\theta=tan^{-1}(0.84)=40^{\circ}$
The angle that the initial velocity makes with respect to the horizontal = $40^{\circ}$
