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Trajectory problem

A certain airplane has a speed of 250.0 km/h and is diving at an angle of 30.0° below the horizontal when the pilot releases a radar decoy. The horizontal distance between the release point and the point where the decoy strikes the ground is 700 m. (Neglect air resistance.)
How high was the plane when the decoy was released and how long was the decoy in the air?


First I fount the velocity components:
v(0,x)=250cos(-30)=216.5 m/s
v(0,y)=250sin(-30)=-125 m/s

Then I used this equation:
y=tan(theta)x-(gx^2)/(2(v(0)cos(theta))^2)
I got 455.36 meters, but I'm told that it is wrong. What am I doing incorrectly?
 
The 250.0 velocity is in km/hr. You must first convert this to m/s.
 

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