he drawing shows two planes each dropping an empty fuel tank. At the moment of release each plane has the same speed of 185 m/s, and each tank is at the same height of 3.62 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.
Well, well, well! Looks like we have a case of gravity and projectile motion here! Let's get to it, shall we?
(a) To find the magnitude of the velocity at which the fuel tank from plane A hits the ground, we need to break it down into its horizontal and vertical components. Since the tank is dropped at an angle of 15.0° above the horizontal, we can use trigonometry to find these components.
The horizontal component of the velocity remains the same as the initial speed of the plane, which is 185 m/s.
For the vertical component, we can use the formula Vf = Vi + at. The initial velocity (Vi) in the vertical direction is 0 m/s, as the tank is dropped from rest. The acceleration (a) due to gravity is approximately 9.8 m/s^2. We can find the time taken (t) for the tank to hit the ground using the formula t = sqrt(2h/g), where h is the height (3.62 km = 3620 m).
Using these values, we find the time taken to be approximately 27.65 seconds.
Now, we can find the vertical component of the velocity using the formula Vf = Vi + at. Plugging in the values, we get Vf = 0 + (9.8 m/s^2)(27.65 s) = 270.07 m/s.
Using the Pythagorean theorem, we can find the magnitude of the velocity: Magnitude = sqrt((185 m/s)^2 + (270.07 m/s)^2) ≈ 333.97 m/s.
(b) To find the direction, we can use trigonometry once again. The angle can be found using the inverse tangent function. The direction would be given as a positive angle with respect to the horizontal. In this case, we use the inverse tangent of the vertical component divided by the horizontal component. So, the direction is tan^(-1)(270.07 m/s / 185 m/s) ≈ 54.73° above the horizontal.
And now, let's move on to plane B!
(c) Although the speed remains the same, the direction is now 15.0° below the horizontal. This means the vertical component of the velocity will have a negative value. Using similar calculations as before, we find the magnitude to be approximately 333.97 m/s.
(d) Now, let's calculate the direction. The vertical component is negative, so we'll use the formula tan^(-1)(-270.07 m/s / 185 m/s). This gives us an angle of approximately -54.73° below the horizontal.
So, to summarize:
(a) The magnitude of the velocity when the fuel tank is from plane A is approximately 333.97 m/s, and it hits the ground at approximately 54.73° above the horizontal.
(c) The magnitude of the velocity when the fuel tank is from plane B is also approximately 333.97 m/s, but it hits the ground at approximately 54.73° below the horizontal.
Keep those planes and fuel tanks flying, and remember, gravity can be a real downer sometimes!
To solve this problem, we can break down the motion of the fuel tanks into horizontal and vertical components.
(a) to find the magnitude of the velocity with which the fuel tank from plane A hits the ground, we need to calculate its horizontal and vertical components separately.
Given that plane A is flying at an angle of 15.0° above the horizontal, we can calculate its horizontal component as follows:
Horizontal component of velocity = velocity * cos(angle)
Horizontal component of velocity (plane A) = 185 m/s * cos(15°)
Now, let's calculate the vertical component of velocity for plane A:
Vertical component of velocity = velocity * sin(angle)
Vertical component of velocity (plane A) = 185 m/s * sin(15°)
The magnitude of the velocity can be found using the Pythagorean theorem:
Magnitude of velocity (plane A) = sqrt(horizontal component^2 + vertical component^2)
(b) The direction of the velocity will be the angle formed with the horizontal axis.
Direction of the velocity (plane A) = atan(vertical component / horizontal component)
Now, let's move on to finding the magnitude and direction of the velocity for the fuel tank from plane B.
(c) Plane B is flying at an angle of 15.0° below the horizontal. We can use the same process as before to find its magnitude and direction.
Horizontal component of velocity (plane B) = 185 m/s * cos(15°)
Vertical component of velocity (plane B) = 185 m/s * sin(-15°)
Magnitude of velocity (plane B) = sqrt(horizontal component^2 + vertical component^2)
Direction of the velocity (plane B) = atan(vertical component / horizontal component)
Note: In this case, since the angle is below the horizontal, we use the negative value of the angle when calculating the vertical component.
Now, you can substitute the values into the equations to find the specific values for each part of the question.
To solve this problem, we need to separate the motion of the fuel tank into horizontal and vertical components. Let's start by finding the horizontal and vertical velocities of the fuel tank released by plane A.
(a) Magnitude of velocity (vA):
The vertical component of the velocity is given by the initial speed (185 m/s) multiplied by the sine of the angle (15°). So, the vertical velocity (vAy) is:
vAy = 185 m/s * sin(15°).
The horizontal component of the velocity is given by the initial speed (185 m/s) multiplied by the cosine of the angle (15°). So, the horizontal velocity (vAx) is:
vAx = 185 m/s * cos(15°).
Using these two components, we can calculate the magnitude of the velocity (vA) using the Pythagorean theorem:
vA = √(vAx^2 + vAy^2).
(b) Direction of velocity (θA):
The direction of the velocity is given by the inverse tangent of the vertical velocity divided by the horizontal velocity:
θA = atan(vAy / vAx).
Now, let's find the horizontal and vertical velocities of the fuel tank released by plane B.
(c) Magnitude of velocity (vB):
The vertical component of the velocity is the same as for plane A, so vBy = vAy. The horizontal component, however, is now negative because the angle is below the horizontal. So, the horizontal velocity (vBx) is:
vBx = -185 m/s * cos(15°).
Again, using the Pythagorean theorem, we can calculate the magnitude of the velocity (vB):
vB = √(vBx^2 + vBy^2).
(d) Direction of velocity (θB):
The direction of the velocity is given by the inverse tangent of the vertical velocity divided by the absolute value of the horizontal velocity:
θB = atan(vBy / |vBx|).
Now, let's substitute the values and calculate the answers:
vAy = 185 m/s * sin(15°).
vAx = 185 m/s * cos(15°).
vA = √(vAx^2 + vAy^2).
θA = atan(vAy / vAx).
vBy = vAy.
vBx = -185 m/s * cos(15°).
vB = √(vBx^2 + vBy^2).
θB = atan(vBy / |vBx|).
Solving these equations will give us the answers to the parts (a), (b), (c), and (d) of the problem.
I'll do (a) and (b). The other plane is similar.
Since the plane is flying at an angle of 15° upward, the tank when released as a velocity of
185sin15° = 47.88 m/s vertically
185cos15° = 178.70 m/s horizontally
How long does it take the tank to hit the ground?
3620+47.88t-4.9t^2 = 0
t = 32.50
So, the vertical speed on impact is
47.88-32.50*9.8 = -270.62
Thus, on impact, the speed is
√(178.70^2 + 270.62^2) = 324.30 m/s
In a direction θ where
tanθ = -270.62/178.70 = -56.56°