As preparation for this problem, review Conceptual Example 10. The drawing shows two planes each dropping an empty fuel tank. At the moment of release each plane has the same speed of 105 m/s, and each tank is at the same height of 4.79 km 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 15.0° above the horizontal and the other is flying at an angle of 15.0° below the horizontal. Find the (a) magnitude and (b) direction of the velocity with which the fuel tank hits the ground if it is from plane A. Find the (c) magnitude and (d) direction of the velocity with which the fuel tank hits the ground if it is from plane B. In each part, give the direction as a positive angle with respect to the horizontal.
To solve this problem, we need to break it down into different components and use the concepts of projectile motion. Let's go step by step.
Step 1: Analyzing the motion of the fuel tank from Plane A and Plane B.
We are given that both planes have the same speed of 105 m/s but are flying in different directions. Plane A is flying at an angle of 15.0° above the horizontal, while Plane B is flying at an angle of 15.0° below the horizontal.
Step 2: Breaking down the velocities into horizontal and vertical components.
The velocity of an object can be broken down into its horizontal (Vx) and vertical (Vy) components using trigonometry. We can use the following equations:
Vx = V * cos(θ)
Vy = V * sin(θ)
Where V is the magnitude of the velocity and θ is the angle with respect to the horizontal.
Step 3: Finding the components of velocity for each plane.
For Plane A:
Vx_A = 105 * cos(15.0°)
Vy_A = 105 * sin(15.0°)
For Plane B:
Vx_B = 105 * cos(-15.0°) (since it is flying below the horizontal)
Vy_B = 105 * sin(-15.0°) (since it is flying below the horizontal)
Note: The negative sign in front of the angle indicates that the motion is in the opposite direction.
Step 4: Velocity of the fuel tank when it hits the ground.
For both planes, the fuel tank will experience free fall motion in the vertical direction (Vy) and a constant horizontal velocity (Vx).
(a) Magnitude of velocity when the fuel tank hits the ground from Plane A:
The magnitude of the velocity (V) can be found using the Pythagorean theorem:
V_A = √(Vx_A^2 + Vy_A^2)
(b) Direction of velocity when the fuel tank hits the ground from Plane A:
The direction can be found using the inverse tangent function:
θ_A = tan^(-1)(Vy_A / Vx_A)
(c) Magnitude of velocity when the fuel tank hits the ground from Plane B:
V_B = √(Vx_B^2 + Vy_B^2)
(d) Direction of velocity when the fuel tank hits the ground from Plane B:
θ_B = tan^(-1)(Vy_B / Vx_B)
Step 5: Calculate the values.
Using the given values and the equations above, plug in the values and calculate the results:
(a) V_A = √((105 * cos(15.0°))^2 + (105 * sin(15.0°))^2)
(b) θ_A = tan^(-1)((105 * sin(15.0°)) / (105 * cos(15.0°)))
(c) V_B = √((105 * cos(-15.0°))^2 + (105 * sin(-15.0°))^2)
(d) θ_B = tan^(-1)((105 * sin(-15.0°)) / (105 * cos(-15.0°)))
Step 6: Calculate the values using a calculator.
Using a calculator, input the given values into the equations above and calculate the results to find the magnitude and direction of the velocity when the fuel tanks hit the ground for both Plane A and Plane B.
Once you have calculated these values, you will have the (a) magnitude and (b) direction of the velocity when the fuel tank from Plane A hits the ground, as well as the (c) magnitude and (d) direction of the velocity when the fuel tank from Plane B hits the ground.