physics
posted by brenda .
A shotputter throws the shot with an initial speed of 15.5 m/s at a 32.0 degree angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.20 m above the ground.

Split the initial velocity (u) into two components, vertical (uy) and horizontal (ux) using:
ux=u cos(θ) and
uy=u sin(θ)
Solving the time (t) the shot stays in the air from:
Sy=uy*t  (1/2)gt²
where Sy = vertical distance travelled = 2.2m
(reject the negative root).
The horizontal distance travelled is then
Sx=ux*t 
What do you mean by where Sy = vertical distance traveled = 2.2m? Is the 2.2m from the equation above? Do I add that in to the equation?

2.2m is the distance between the hand and the ground.
So we equate the vertical distance (from the hand) with the distance as a function of time:
2.2=uy*t  (1/2)gtĀ²
Solve for t. 
24.1 m