physics
posted by someone on .
a shotputter throws the shot with an intial speed of 14 m/s at a 40 degree angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.2m above the ground.
i got to answers which is corret?
47.17m or 57.24m or neither?

What is the vertical velocity component?
how long will it stay in the air?
hfinal=hinitial+Vv*timeinair1/2 g timinair^2
solve for time in air.
Then horizontal distance=Vh*timeinair
I didn't get your answers. 
What is the vertical velocity component?
that's 9 m/s right?
the time i got is 4.4/5.34s
which i think are wrong.. 
correct. I didn't check the quadreatic solutions, but it is slighly over 1.8 seconds

so which should i use 4.4 or 5.34?
which is slighty over 1.8seconds anyway? and shud i just subtract it? 
Neither of your answers is correct. The shot returns to 2.2 m height after travelling
X = (V^2/g)sin(2*40) = 19.7 m horizontally. It will only travel another few meters before hitting the ground, about 1/4 second later. BobP's quadratic solution will give the exact value 
so the quadratic solution is like this 4.9t^2+9t+2.2=0?

Isn't there a negative sign before the 4.9?

okay so t=2.06?
so x=22.08?