An airplane is flying with a velocity of 240 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altitude of the plane is 2.2 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 analyze the given information and break it down into components. Let's start by finding the horizontal and vertical components of the airplane's velocity.
Given:
Velocity magnitude, v = 240 m/s
Angle with the horizontal, θ1 = 30.0°
Altitude of the plane, h = 2.2 km = 2200 m
First, let's find the horizontal component of the velocity (v_x). We can use trigonometry to do this.
v_x = v * cos(θ1)
Substituting the given values:
v_x = 240 m/s * cos(30.0°)
v_x = 240 m/s * 0.866
v_x ≈ 207.84 m/s
Next, let's find the time it takes for the flare to hit the ground. Since there is no vertical acceleration (assuming no air resistance), the time taken to reach the ground will be the same as the time taken by the airplane to travel horizontally to the target.
Let's denote the time taken as t.
Distance traveled horizontally, d = v_x * t
Since the flare is released at a height of 2.2 km (2200 m), the total distance traveled vertically is h.
Using the kinematic equation for vertical motion:
h = (1/2) * g * t^2
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearranging the equation:
t^2 = (2 * h) / g
t = sqrt((2 * h) / g)
Substituting the given values:
t ≈ sqrt((2 * 2200 m) / 9.8 m/s^2)
t ≈ sqrt(4400 / 9.8) ≈ sqrt(448.98) ≈ 21.2 s
Now that we have the time taken, we can find the distance traveled horizontally.
d = v_x * t
d ≈ 207.84 m/s * 21.2 s
d ≈ 4406.85 m ≈ 4.41 km
Finally, to find the angle θ, we can use trigonometry again.
θ = tan^(-1)(h/d)
Substituting the given values:
θ ≈ tan^(-1)(2200 m / 4406.85 m)
θ ≈ tan^(-1)(0.499)
θ ≈ 26.9°
Therefore, the angle θ is approximately 26.9°.