Question

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 40t-16(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

Answer 1

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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.