Solve Trajectory Problem: Plane Height & Decoy Flight Time

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The airplane is diving at an angle of 30° below the horizontal and has a speed of 250.0 km/h, which needs to be converted to meters per second for accurate calculations. The horizontal distance from the release point to the impact point of the decoy is 700 m. The initial velocity components were calculated as 216.5 m/s horizontally and -125 m/s vertically. The user applied the trajectory equation but received incorrect results, indicating a potential error in unit conversion or calculation. Correctly converting the speed to m/s and re-evaluating the calculations will yield the accurate height of the plane and the decoy's flight time.
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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?
 
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The 250.0 velocity is in km/hr. You must first convert this to m/s.
 
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