Richard P. answered 10/14/18
PhD in Physics with 10+ years tutoring experience in STEM subjects
A key to many problems of this type is to think about the time-of-flight (TOF). That is the amount of time that the object is in free fall. For the vertical part of this problem we have
5620 = (1/2) g TOF^2
Plugging in g = 9.8 then solving for TOF gives TOF = 33.9 s.
The horizontal component of the velocity never changes (so is equal to 110 m/s), the vertical component is given by v = g TOF = 331.9 m/s
The magnitude (overall) velocity at impart is given by the the square root sum of squares .
That is |v| = sqrt ( 110^2 + 331.9^2) = 349.7m/s