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