Since the projectile is released its initial velocity is the same as the velocity of the plane at the time of release

Take the origin at the point of release

Let $x$ and $y(=-\mathrm{730 }\mathrm{m})$ be the coordinates of the point on the ground where the projectile hits and let $t$ be the time when it hits. Then

$y=-v_{0}t\mathrm{cos\theta }-\frac{1}{2}\mathrm{g}{t}^2$

where θ$=53.0°$

This equation gives

$v_0=-\frac{y+\frac{1}{2}\mathrm{g}{t}^2}{t\mathrm{cos\theta }}$

$=-\frac{-730+\frac{1}{2}\left(9.8\right){\left(5\right)}^2}{5\cos 53°}=202 \mathrm{m }{\mathrm{s}}^{-1}$