An airplane is flying with a velocity of 248 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altitude of the plane is 2.5 km, a flare is released from the plane. The flare hits the target on the ground. What is the angle θ?

To find the angle θ, we need to break down the given information into its components and use trigonometry to solve for the angle.

First, let's find the horizontal and vertical components of the airplane's velocity. We can use the given velocity (248 m/s) and the angle (30.0°) to find these components.

The horizontal component (Vx) can be calculated using the formula:
Vx = V * cos(θ)
where V is the velocity and θ is the angle.

Substituting the given values:
Vx = 248 m/s * cos(30.0°)
Vx = 248 m/s * 0.866
Vx ≈ 214.9 m/s

The vertical component (Vy) can be calculated using the formula:
Vy = V * sin(θ)
where V is the velocity and θ is the angle.

Substituting the given values:
Vy = 248 m/s * sin(30.0°)
Vy = 248 m/s * 0.5
Vy = 124 m/s

Now, let's consider the motion of the flare after it is released from the airplane. We know that the flare's vertical velocity is the same as the airplane's (Vy = 124 m/s). The horizontal velocity of the flare is 0 because there is no force acting on it horizontally once it is released.

To find the angle θ, we can use the tangent function:

tan(θ) = Vy / Vx

Substituting the values we found:
tan(θ) = 124 m/s / 214.9 m/s

Now, let's solve for θ:

θ = tan^-1(124 m/s / 214.9 m/s)

Using a calculator, we find that θ is approximately 30.2°.

Therefore, the angle θ is approximately 30.2°.