Solve Trajectory Problem: Plane Height & Decoy Flight Time

[uberbandgeek6]

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?

 

[Mattowander]

The 250.0 velocity is in km/hr. You must first convert this to m/s.

 

[baba89]

There are a few potential issues with your calculation. First, make sure that you are using the correct units for your calculations. The given speed of 250.0 km/h should be converted to meters per second (m/s) for consistency. Also, when finding the horizontal and vertical velocity components, make sure to use the correct angle of 30 degrees (not -30 degrees).

Another potential issue could be with your use of the equation for trajectory. It is important to note that this equation assumes a projectile motion with a constant velocity, which may not accurately describe the motion of the decoy in this situation. Additionally, the equation you have used appears to be missing a term for the initial height of the projectile, which may be a factor in your incorrect result.

To accurately solve this problem, you may need to use a more sophisticated approach, such as using the equations of motion and kinematics to analyze the motion of the decoy. Alternatively, you could try using a physics simulation tool to model the situation and obtain more accurate results.