Answer
The speed of the projectile must be $68.4~m/s$ when it is fired from the cannon.
Work Step by Step
We can find an expression for the time of flight:
$t = \frac{x}{v_x} = \frac{x}{v_0~cos~\theta}$
We can find the initial speed of the projectile:
$y = v_{0y}t+\frac{1}{2}a_yt^2$
$y = v_0~sin~\theta~(\frac{x}{v_0~cos~\theta})+\frac{1}{2}a_y(\frac{x}{v_0~cos~\theta})^2$
$y = x~tan~\theta+\frac{1}{2}a_y(\frac{x}{v_0~cos~\theta})^2$
$y - x~tan~\theta = \frac{a_y~x^2}{2~(v_0~cos~\theta)^2}$
$v_0^2 = \frac{a_y~x^2}{2~(cos~\theta)^2~( y - x~tan~\theta)}$
$v_0 = \sqrt{\frac{a_y~x^2}{2~(cos~\theta)^2~( y - x~tan~\theta)}}$
$v_0 = \sqrt{\frac{(-9.80~m/s^2)(350~m)^2}{2~(cos~40.0^{\circ})^2~(75.0~m - (350~m)~tan~40.0^{\circ})}}$
$v_0 = 68.4~m/s$
The speed of the projectile must be $68.4~m/s$ when it is fired from the cannon.
