Answer
$v_0 \approx 46.6 ft/second$
Work Step by Step
When $y_0 \ne 0$, then we have: $y-y_0=-\dfrac{g}{2v_{0}^2\cos^2 \alpha }x^2+\tan \alpha x$
Plug in the data:
$0-6.5 =-\dfrac{g}{2v_{0}^2\cos^2 40^{\circ} }(73 \ ft 10 \ in )^2+\tan 40^{\circ} (73 \ ft 10 \ in )^2$
Since,$1 \ ft =1 \ in ; g= 32 ft/sec^2$
So, $-6.5 =\dfrac{-32}{2v_{0}^2\cos^2 40^{\circ} } (73.833)^2+\tan 40^{\circ} (73.833)$
or, $-6.5 \approx \dfrac{-148633.6}{v_{0^2}}+61.95$
or, $v_0 \approx \sqrt {\dfrac{148633.6}{68.45}}$
or, $v_0 \approx 46.6 ft/second$
