Since the flare is released from the plane, it has the same initial velocity as the plane, i.e., 240 m/s at an angle of 30 degrees with the horizontal.
Let's assume that the distance from the plane to the target is d, and the altitude of the plane is h. The angle θ is the angle between the plane and the target.
Using trigonometry, we can write:
tan θ = h/d
We need to find the value of θ. To do that, we need to find the values of h and d.
We know that the altitude of the plane is 2.4 km = 2400 m. Let's call the time it takes for the flare to hit the target t. Since the flare is moving under gravity, its motion can be described as:
h = (1/2)gt^2
where g is the acceleration due to gravity, which is approximately 9.81 m/s^2.
The horizontal distance traveled by the flare in time t is:
d = vt
where v is the horizontal component of the velocity of the flare/plane, which is given by:
v = 240 cos 30° = 240 × √3/2 = 120√3 m/s
Equating the expressions for h and d, we get:
(1/2)gt^2 = vt
Solving for t, we get:
t = 2h/g = 2 × 2400/9.81 = 489.55 s
Substituting this value of t in the expression for d, we get:
d = vt = 120√3 × 489.55 ≈ 70000 m
Now we can find θ:
tan θ = h/d = 2400/70000 ≈ 0.0343
θ = tan^(-1)(0.0343) ≈ 1.96°
Therefore, the angle θ is 1.96 degrees.