Precalculus
posted by Kate .
A discus is thrown from a height of 3 feet with an initial velocity of 55 ft/s at an angle of 44° with the horizontal. How long will it take for the discus to reach the ground?
I started off with 55cos(44)=40
then I made an equation
40t16(t)=2.5 seconds?
I feel like I did this wrong because I didn't add 3 feet anywhere
or do I add 3 in to 55cos(44) giving me an answer of 2.6

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.