Posted by brenda on .
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.

physics 
MathMate,
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 
physics 
Sam,
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?

physics 
MathMate,
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. 
physics 
self,
24.1 m