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Yes, the 3 feet are important.
You can consider the datum (y=0) at 3' from the ground, and the ground is therefore at -3 feet.
The vertical initial velocity, vy, is u*sin(θ), so it gives
vy=55sin(44°)=38.21 approx.
The (vertical) distance travelled is given by:
S=vy*t-(1/2)gt²
where
S=-3 (ground)
vy=55sin(44°), and
g=32.2 ft/s²
Solve for t.
I get -0.08 and 2.45 s.
