1, An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.
A) Express the initial velocity of the package as a vector. What assumptions have made about the frame of reference? you
B) How long will it take for the package to reach the ground?
C) How fast will it be going as it lands? Express your answer as a vector.
D) Describe the path of the package as seen by a stationary
observer on the ground.
E) Describe the path of the package as seen by someone in the aeroplane.
A) The initial velocity of the package can be expressed as a vector of (70 m/s, 0 m/s). We have assumed that the frame of reference is the ground.
B) It will take approximately 11.43 seconds for the package to reach the ground.
C) The package will be going at a velocity of (70 m/s, -9.8 m/s) as it lands.
D) From the perspective of a stationary observer on the ground, the package will appear to move in a straight line from the aircraft to the ground.
E) From the perspective of someone in the aeroplane, the package will appear to move in a curved path due to the aircraft's motion.
A) The initial velocity of the package can be expressed as a vector with a magnitude of 70 m/s and a direction of eastwards. The assumption made about the frame of reference is that the aircraft's velocity is the same as the package's initial velocity.
B) To determine the time it takes for the package to reach the ground, we need to know the acceleration due to gravity. Assuming standard gravity (9.8 m/s^2), we can use the kinematic equation:
h = ut + (1/2)at^2
where:
h = height (800 m)
u = initial velocity (0 m/s, since the package is dropped from rest)
a = acceleration due to gravity (-9.8 m/s^2, considering downward direction)
t = time taken
Rearranging the equation and solving for t, we get:
800 = 0 + (1/2)(-9.8)t^2
1600 = -4.9t^2
t^2 = 1600 / 4.9
t ≈ 8.15 seconds
Therefore, it will take approximately 8.15 seconds for the package to reach the ground.
C) To determine the final velocity, we can use another kinematic equation:
v = u + at
where:
v = final velocity
u = initial velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time taken (8.15 seconds)
Plugging in the values, we get:
v = 0 + (-9.8)(8.15)
v ≈ -79.87 m/s
Expressed as a vector, the final velocity of the package as it lands will be approximately -79.87 m/s in the downward direction.
D) As seen by a stationary observer on the ground, the path of the package will be a vertical straight line. It will fall straight downwards from the aircraft to the ground due to the force of gravity.
E) As seen by someone in the aeroplane, the path of the package will appear to be a parabolic trajectory. This is because both the package and the plane are moving horizontally at the same velocity. However, due to the force of gravity acting vertically, the package will follow a curved path and eventually land on the ground below the plane.
A) To express the initial velocity of the package as a vector, we need to consider both the magnitude (speed) and direction. The magnitude remains the same as that of the aircraft, which is 70 m/s. Since the aircraft is flying eastwards, the direction of the package's velocity vector will also be eastward. Therefore, the initial velocity of the package can be expressed as a vector: 70 m/s eastward.
The assumption made about the frame of reference is that we are considering a stationary observer on the ground. This means that we are observing the aircraft and the package's motion from the perspective of someone who is not moving relative to the ground.
B) To determine how long it will take for the package to reach the ground, we need to calculate the time it takes for the package to fall from a height of 800 m. We can use the equation for vertical motion under constant acceleration:
h = (1/2) * g * t^2
where h is the initial height (800 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Solving the equation for time, we get:
t^2 = (2h) / g
t = sqrt((2 * 800) / 9.8)
t ≈ 12.24 s
Therefore, it will take approximately 12.24 seconds for the package to reach the ground.
C) To determine the speed of the package as it lands, we need to know the final velocity just before it reaches the ground. Assuming no additional forces act on the package during its fall, its final velocity will be solely determined by the acceleration due to gravity.
The final velocity can be calculated using the equation:
v = u + gt
where v is the final velocity, u is the initial velocity (which is 0 m/s since the package was dropped), g is the acceleration due to gravity, and t is the time taken to reach the ground (which we calculated to be approximately 12.24 s).
Plugging in the values, we get:
v = 0 + (9.8 * 12.24)
v ≈ 119.95 m/s
Therefore, the speed of the package as it lands is approximately 119.95 m/s.
Since the package was dropped vertically downwards, the direction of its velocity vector can be considered as downward or "towards the ground."
D) As seen by a stationary observer on the ground, the path of the package will be a straight vertical line downward. The package will appear to fall vertically from the perspective of the observer.
E) From the perspective of someone in the airplane, the path of the package will also appear to be a straight vertical line downward. Since the airplane is traveling at a steady horizontal velocity, the package will appear to fall straight down relative to the aircraft.