Projectile motion of shotputter

1. The problem statement, all variables and given/known data
A shotputter throws a shot with an initial speed of 15.5m/s at a 34.0 degree angle to the horizontal. Calculate the horizontal distance given that the shot leaves the athletes hand 2.2m above the ground???


2. Relevant equations
Distance=((initial velocity)^2 x sin (2xdegrees))/gravity(9.8)

Although this is the level range formula used when height final = height of initial. How do you change this formula to accommodate the fact that the shot lands 2.2m lower than it started????????

3. The attempt at a solution
d=15.5^2 x sin 68/ 9.8 =29.09m, I know this must be wrong but i have read through the text book and it does not help me at all in finding how to work this out. I shows what to do if the degrees is not given but with the degrees what equation can i use??????????????

 

Yeah but none of them incorporate the angle of release given???? Im guessing thats an important variable, so shouldnt it be used

 

Your equations must be wrong
0=2.2+15.5*sin 34*t-1/2*9.8*t^2
t=-0.225
x=15.5*cos 34*-0.225
x=-2.8912
this answer does not make sense

 

alright, cheers