Two planes are about to drop an empty fuel tank. At the moment of release each plane has the same speed of 135 m/s, and each tank is at the same height of around 2.00km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15deg above the horizontal (Plane A) and the other is flying at an angle of 15deg below the horizontal (Plane B). Find the magnitude and the direction of the velocity with which the fuel tank hits the ground if it is from a) Plane A b) Plane B. Give the directional angles with respect to the horizontal.
a) The velocity of the fuel tank dropped from Plane A can be found by resolving the initial velocity into horizontal and vertical components.
The horizontal component of velocity can be found using the formula: v = v₀ * cos(θ), where v₀ is the initial velocity (135 m/s) and θ is the angle of 15 degrees above the horizontal.
v_horizontal = 135 m/s * cos(15 degrees) ≈ 130.84 m/s
The vertical component of velocity can be found using the formula: v = v₀ * sin(θ).
v_vertical = 135 m/s * sin(15 degrees) ≈ 36.92 m/s
The magnitude of the velocity with which the fuel tank hits the ground can be found using the Pythagorean theorem: v = √(v_horizontal² + v_vertical²).
v = √(130.84 m/s)² + (36.92 m/s)² ≈ 137.09 m/s (approximately)
To find the direction of the velocity, use the tangent inverse (arctan) function: θ = arctan(v_vertical / v_horizontal).
θ = arctan(36.92 m/s / 130.84 m/s) ≈ 16.25 degrees (approximately)
Therefore, the magnitude of the velocity is approximately 137.09 m/s and the direction angle, with respect to the horizontal, is approximately 16.25 degrees.
b) The same process can be applied to find the magnitude and direction of the velocity with which the fuel tank hits the ground if it is dropped from Plane B.
The horizontal component of velocity remains the same: 130.84 m/s.
However, the vertical component of velocity will be negative due to the angle of 15 degrees below the horizontal. So,
v_vertical = -36.92 m/s
The magnitude of the velocity can be found using the Pythagorean theorem again:
v = √(130.84 m/s)² + (-36.92 m/s)² ≈ 135.97 m/s (approximately)
To find the direction, use the arctan function again, taking into account the negative value of the vertical component:
θ = arctan((-36.92 m/s) / 130.84 m/s) ≈ -16.25 degrees (approximately)
Therefore, the magnitude of the velocity is approximately 135.97 m/s, and the direction angle, with respect to the horizontal, is approximately -16.25 degrees.
To find the magnitude and direction of the velocity with which the fuel tank hits the ground, we can start by analyzing the vertical and horizontal components of the velocities for each plane.
Let's break down the given information:
Initial speed of each plane (v): 135 m/s
Height of release (h): 2.00 km (which is 2000 m)
Angle of Plane A above the horizontal (θA): 15°
Angle of Plane B below the horizontal (θB): -15°
Step 1: Calculate the horizontal and vertical components of the velocity for each plane.
For Plane A:
Horizontal component (vAx) = v * cos(θA)
Vertical component (vAy) = v * sin(θA)
For Plane B:
Horizontal component (vBx) = v * cos(θB)
Vertical component (vBy) = v * sin(θB)
Step 2: Determine the time taken for the fuel tank to hit the ground.
The time taken for an object to fall freely under gravity can be calculated using the equation:
h = 0.5 * g * t^2
where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Solving for t, we get:
t = √(2h / g)
Step 3: Calculate the final velocity of the fuel tank when it hits the ground.
The final velocity of the fuel tank can be determined by multiplying the time taken with the acceleration due to gravity:
Final velocity = t * g
Step 4: Determine the magnitude and direction of the velocity.
The magnitude of the velocity is the final velocity obtained in Step 3.
For Plane A:
Magnitude of velocity (VA) = Final velocity of Plane A = t * g = √(2h / g) * g
For Plane B:
Magnitude of velocity (VB) = Final velocity of Plane B = t * g = √(2h / g) * g
The direction of the velocity can be given in terms of the angle with respect to the horizontal.
For Plane A:
Directional angle with respect to the horizontal (θVA) = θA
For Plane B:
Directional angle with respect to the horizontal (θVB) = - θB
Substituting the given values, we can calculate the magnitude and direction of the velocity for each case.
Magnitude and Direction of Velocity:
a) Plane A:
Magnitude of velocity (VA) = √(2 * 2000 / 9.8) * 9.8 ≈ 630.83 m/s
Directional angle with respect to the horizontal (θVA) = 15° (above the horizontal)
b) Plane B:
Magnitude of velocity (VB) = √(2 * 2000 / 9.8) * 9.8 ≈ 630.83 m/s
Directional angle with respect to the horizontal (θVB) = -15° (below the horizontal)
Therefore, the magnitude and direction of the velocity with which the fuel tank hits the ground from Plane A is approximately 630.83 m/s at 15° above the horizontal, and from Plane B is also approximately 630.83 m/s at 15° below the horizontal.
A
B
A
D
B
v(ox) =v(o) •cosα= 135 •cos15 = 130.4 m/s,
v(oy) =v(o) •sinα= 135 •sin15 = 34.9 m/s,
PlaneA:
the tank moves upward along the parabolic path until its v(y) =0.
The height hₒ= v²(o) •sin²α/2•g =135² •sin²15/2•9.8 = 62.3 m
H=h+hₒ=2000+62.3 =2062.3 m.
t =sqrt(2H/g) = sqrt(2•2062.3/9.8) = 20.5.
The vertical component of the velocity near the ground is
v(y)=gt = 9.8•20.5 = 201.05 m/s.
v =sqrt(v²(x)+v²(y)) = sqrt(130.4² +201.05²) = 239.6 m/s.
tanφ =v(y)/v(x) =201.05/130.4 =1.54,
φ=57º
Plane B.
h=gt²/2 => t=sqrt(2h/g) =sqrt(2•2000/9.8) = 20.2 s.
v(y) = v(oy)+gt = 34.9+9.8•20.2 = 232.9 m/s.
v =sqrt(v²(x)+v²(y)) = sqrt(130.4² +232.9²) = 266.9 m/s.
tanφ =232.9/130.4 = 1.789,
φ = 60.8º.