Answer
The ball will leave the gun at a speed of 9.98 m/s
Work Step by Step
We can find the spring constant $k$ of the spring as:
$kx = F$
$k = \frac{F}{x}$
$k = \frac{91.0~N}{0.175~m}$
$k = 520~N/m$
We then find the speed $v_{max}$ when the ball leaves the gun;
$v_{max} = \sqrt{\frac{k}{m}}~A$
$v_{max} = \sqrt{\frac{520~N/m}{0.160~kg}}~(0.175~m)$
$v_{max} = 9.98~m/s$
Therefore, the ball will leave the gun at a speed of 9.98 m/s.
