Answer
$v_{\circ}=20m/s$
Work Step by Step
We know that
$x=x_{\circ}+(v_{\circ}cos\theta) t$
We plug in the known values to obtain:
$41m=(v_{\circ}cos40^{\circ})t$
$\implies t=\frac{53.52m}{v_{\circ}}$
Now $y=y_{\circ}+(v_{\circ}sin\theta) t-\frac{1}{2}gt^2$
We plug in the known values to obtain:
$0=0.75m+v_{\circ}(0.6428)(\frac{53.52m}{v_{\circ}})-\frac{1}{2}(9.8m/s^2)(\frac{53.52m}{v_{\circ}})^2$
This simplifies to:
$v_{\circ}=20m/s$
