The drawing shows two planes each dropping an empty fuel tank. At the moment of release each plane has the same speed of 173 m/s, and each tank is at the same height of 2.75 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
horizontal speed is the same for both: vx = 173 cos15°
for each, the height of the tank from dropping is
h(t) = 2750 ± 173sin15° t - 4.9t^2
so, find t when h=0
then the y component is vy = -9.8t
landing speed is s^2 = vx^2 + vy^2
landing angle is tanθ = vy/vx
Oops. the y component is vy = ±173sin15° - 9.8t
To solve this problem, we will break it down into components and use the equations of motion for projectiles.
Given:
- Initial speed of both planes = 173 m/s
- Height of release for both tanks = 2.75 km
- Plane A is flying at an angle of 15.0° above the horizontal
- Plane B is flying at an angle of 15.0° below the horizontal
a) To find the magnitude of the velocity with which the fuel tank hits the ground if it is from plane A:
Step 1: Determine the initial velocity components for plane A.
We can use trigonometry to find the vertical and horizontal components of velocity.
Vertical (Vy): Vy = V_initial * sin(angle)
Horizontal (Vx): Vx = V_initial * cos(angle)
Step 2: Calculate the time of flight for the tank from plane A.
We know that the height of release is 2.75 km (or 2750 m), and we assume the free-fall acceleration to be approximately 9.8 m/s^2.
Using the equation: height = (1/2) * g * t^2 (where g is the acceleration due to gravity and t is the time of flight), we can solve for t.
2.75 km = (1/2) * 9.8 m/s^2 * t^2
5500 = 4.9 t^2
t^2 = 5500 / 4.9
t ≈ 32.26 s (rounded to 2 decimal places)
Step 3: Calculate the vertical component of velocity when the tank hits the ground.
Using the equation: Vy_final = Vy_initial + g * t, we can substitute the values to calculate Vy_final.
Vy_final = 0 + 9.8 m/s^2 * 32.26 s
Vy_final ≈ 316.35 m/s (rounded to 2 decimal places)
b) To find the direction of the velocity with which the fuel tank hits the ground if it is from plane A:
The direction can be found using trigonometry.
The tangent of the angle with respect to the horizontal can be calculated as follows: tan(angle) = Vy_final / Vx
tan(angle) = 316.35 m/s / Vx
To solve for the angle, take the inverse tangent (arctan) of both sides of the equation.
angle = arctan(316.35 m/s / Vx)
c) To find the magnitude of the velocity with which the fuel tank hits the ground if it is from plane B:
Since plane B is flying at an angle of 15.0° below the horizontal, its vertical component of velocity will be negative while the horizontal component remains the same.
Follow the same steps as above to calculate the magnitude and direction, but this time use the negative angle for plane B when finding the direction.
d) To find the direction of the velocity with which the fuel tank hits the ground if it is from plane B:
The direction can be found using the same trigonometric equation as in part b, but with the negative angle for plane B.
angle = arctan(316.35 m/s / Vx) (with the negative angle for plane B)
To solve this problem, we can break down the motion of the fuel tanks into horizontal and vertical components. Let's start by finding the horizontal and vertical velocities of the fuel tanks.
Given:
Initial speed of both planes, v = 173 m/s
Height of tank above the ground, h = 2.75 km = 2750 m
(a) Magnitude of velocity when tank hits the ground from plane A:
Since plane A is flying at an angle of 15.0° above the horizontal, the horizontal component of velocity remains the same as the initial speed: v_Ax = v_A = 173 m/s.
The vertical component of velocity can be found using trigonometry:
v_Ay = v_A * sin(angle above horizontal)
= 173 m/s * sin(15°)
(b) Direction of velocity when tank hits the ground from plane A:
By convention, direction measured above the horizontal is considered positive. So the direction of the velocity can be written as 90° + angle above horizontal.
(c) Magnitude of velocity when tank hits the ground from plane B:
Since plane B is flying at an angle of 15.0° below the horizontal, the horizontal component of velocity remains the same as the initial speed: v_Bx = v_B = 173 m/s.
The vertical component of velocity can also be found using trigonometry:
v_By = v_B * sin(angle below horizontal)
= 173 m/s * sin(15°)
(d) Direction of velocity when tank hits the ground from plane B:
Similarly, the direction can be written as 90° - angle below horizontal.
Now, let's calculate the values:
(a) Magnitude of velocity when tank hits the ground from plane A:
v_Ay = 173 m/s * sin(15°)
≈ 44.60 m/s
(b) Direction of velocity when tank hits the ground from plane A:
Angle above horizontal = 90° + 15°
= 105°
(c) Magnitude of velocity when tank hits the ground from plane B:
v_By = 173 m/s * sin(15°)
≈ 44.60 m/s
(d) Direction of velocity when tank hits the ground from plane B:
Angle below horizontal = 90° - 15°
= 75°
Therefore, the answers are:
(a) The magnitude of the velocity when the fuel tank hits the ground from plane A is approximately 44.60 m/s.
(b) The direction of the velocity when the fuel tank hits the ground from plane A is approximately 105° above the horizontal.
(c) The magnitude of the velocity when the fuel tank hits the ground from plane B is approximately 44.60 m/s.
(d) The direction of the velocity when the fuel tank hits the ground from plane B is approximately 75° below the horizontal.