An airplane is flying with a velocity of 240 m/s at an angle of 30.0o with the horizontal,
as the drawing shows. When the altitude of the plane is 2.4 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 velocity of the airplane into its horizontal and vertical components.
Given:
Velocity of the airplane (V) = 240 m/s
Angle of the velocity with the horizontal (θ1) = 30.0°
Altitude of the plane (h) = 2.4 km
First, let's convert the altitude of the plane from km to meters:
Altitude (h) = 2.4 km * 1000 m/km
h = 2400 m
Now, we can find the horizontal and vertical components of the velocity using trigonometry:
Horizontal component of velocity (Vx) = V * cos(θ1)
Vertical component of velocity (Vy) = V * sin(θ1)
Substituting the values, we get:
Vx = 240 m/s * cos(30.0°)
Vx ≈ 240 m/s * 0.866
Vx ≈ 207.8 m/s
Vy = 240 m/s * sin(30.0°)
Vy ≈ 240 m/s * 0.5
Vy ≈ 120 m/s
Now, we can use the vertical equation of motion to find the time it takes for the flare to hit the ground:
Vertical displacement (h) = Vy * t + 0.5 * g * t^2
Since the flare hits the ground, the vertical displacement (h) is equal to 0:
0 = 120 m/s * t + 0.5 * 9.8 m/s^2 * t^2
0 = t * (120 m/s + 4.9 m/s^2 * t)
Simplifying the equation, we get:
0 = 120 m/s * t + 4.9 m/s^2 * t^2
Now, we can solve this quadratic equation for t.
To find the angle θ, we need to use the concept of projectile motion and resolve the velocity vector of the flare into horizontal and vertical components.
Let's break down the given information:
- The airplane is flying with a velocity of 240 m/s at an angle of 30.0° with the horizontal.
- The altitude of the plane is 2.4 km.
First, let's find the initial velocity components of the flare.
The horizontal component (Vx) can be found using the formula:
Vx = V * cos(θ)
where V is the magnitude of the velocity (240 m/s) and θ is the angle of the velocity vector (30°).
Vx = 240 m/s * cos(30°)
Vx = 240 m/s * √3/2
Vx = 240√3 / 2
Vx = 120√3 m/s
The vertical component (Vy) can be found using the formula:
Vy = V * sin(θ)
where V is the magnitude of the velocity (240 m/s) and θ is the angle of the velocity vector (30°).
Vy = 240 m/s * sin(30°)
Vy = 240 m/s * 1/2
Vy = 120 m/s
Now, let's calculate the time it takes for the flare to hit the target on the ground.
We can use the formula for vertical motion:
h = Vy * t + (1/2) * g * t^2
where h is the altitude of the plane (2.4 km), Vy is the vertical component of velocity (120 m/s), t is the time of flight, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Converting the altitude of the plane to meters:
h = 2.4 km * 1000 m/km
h = 2400 m
2400 m = 120 m/s * t + (1/2) * 9.8 m/s^2 * t^2
Rearranging the equation:
0.5 * 9.8 m/s^2 * t^2 + 120 m/s * t - 2400 m = 0
Now, we can solve this quadratic equation to find the time of flight (t).
Solving this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
a = 0.5 * 9.8 m/s^2 = 4.9 m/s^2
b = 120 m/s
c = -2400 m
t = (-120 m/s ± √(120 m/s)^2 - 4 * 4.9 m/s^2 * (-2400 m)) / (2 * 4.9 m/s^2)
t = (-120 m/s ± √14400 m^2 + 47040 m^2) / 9.8 m/s^2
t = (-120 m/s ± √61440 m^2) / 9.8 m/s^2
t = (-120 m/s ± 248 m) / 9.8 m/s^2
Taking the positive value (time cannot be negative):
t = (128 m) / (9.8 m/s^2)
t ≈ 13.06 s
Now, we can use the time of flight (t) to find the horizontal distance traveled by the flare.
We'll use the formula for horizontal motion:
d = Vx * t
where d is the horizontal distance traveled, Vx is the horizontal component of velocity (120√3 m/s), and t is the time of flight (13.06 s).
d = 120√3 m/s * 13.06 s
d ≈ 1975.087882 m
Finally, we can calculate the angle θ using the horizontal distance (d) and the altitude of the plane (h).
tan(θ) = h / d
θ = tan^(-1)(h / d)
θ = tan^(-1)(2400 m / 1975.087882 m)
θ = tan^(-1)(1.215)
Using a calculator, θ ≈ 50.124°
Therefore, the angle θ is approximately 50.124°.
The vertical and horizontal distance components of the flare are
y = 2400 + 120t - 4.9t^2
x = 207.84t
The flare hits the ground when t=37.54
The tangent of the angle of direction at time t=37.54 is
dy/dx = (120-9.8*37.54)/207.84 = -1.193
So, θ = -50°
I am assuming that the angle in question is the angle at which the flare strikes the ground...