Answer
$14.5\frac{m}{s}$
Work Step by Step
We know that
$R=(v_{\circ}cos\theta )t$
$\implies t=\frac{R}{v_{\circ}cos\theta}$.....eq(1)
We also know that
$y=y_{\circ}+(v_{\circ}sin\theta)t-\frac{1}{2}gt^2$
Replacing 't' by the value in eq(1), we obtain:
$\implies 0=y_{\circ}+(v_{\circ}sin\theta)(\frac{R}{v_{\circ}cos\theta})-\frac{1}{2}g(\frac{R}{v_{\circ}cos\theta})^2$
This simplifies to:
$v_{\circ}^2=\frac{gR^2}{2cos^2\theta(y_{\circ}+Rtan\theta)}$
$\implies v_{\circ}=\sqrt{\frac{gR^2}{2cos^2\theta(y_{\circ}+Rtan\theta)}}$
We plug in the known values to obtain:
$v_{\circ}=\sqrt{\frac{(9.81)(23.12)^2}{2cos^2(42.0^{\circ})(1.83+(23.12)tan42.0^{\circ})}}$
$v_{\circ}=14.5\frac{m}{s}$
