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???
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????????
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??????????????