A plane is flying horizontally with speed
169 m/s at a height 4240 m above the ground,
when a package is dropped from the plane.
The acceleration of gravity is 9.8 m/s
2
.
Neglecting air resistance, when the package
hits the ground, the plane will be
1. directly above the package.
2. behind the package.
3. ahead of the package.
020 (part 2 of 4) 10.0 points
What is the horizontal distance from the release point to the impact point?
Answer in units of m.
021 (part 3 of 4) 10.0 points
A second package is thrown downward from
the plane with a vertical speed v1 = 52 m/s.
What is the magnitude of the total velocity
of the package, at the moment it is thrown as
seen by an observer on the ground?
Answer in units of m/s.
022 (part 4 of 4) 10.0 points
What horizontal distance is traveled by this
package?
Answer in units of m
To answer the given questions, we can use the equations of motion under constant acceleration. Let's break down each question and explain how to find the answers step by step.
1. To determine the position of the plane when the package hits the ground, we need to calculate the time it takes for the package to fall. We can use the equation:
h = (1/2) * g * t^2
Where:
h = height from which the package is dropped (4240 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the equation to solve for t:
t^2 = (2 * h) / g
t = sqrt((2 * h) / g)
Once we have the time, we can calculate the horizontal distance traveled by the plane during this time:
distance = speed * time
2. To find the horizontal distance between the release point and the impact point, we can use the equation of motion:
distance = speed * time
where:
distance = horizontal distance traveled by the plane
speed = horizontal speed of the plane (169 m/s)
time = time calculated in the previous step
3. For this question, we need to find the total velocity of the second package as seen by an observer on the ground. The total velocity will include both the horizontal and vertical components. The horizontal component will be the same as the speed of the plane (169 m/s), and the vertical component can be calculated using the formula:
v = u + gt
where:
v = final velocity in the vertical direction
u = initial velocity in the vertical direction (52 m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = time for the package to reach the ground (calculated in the first question)
The magnitude of the total velocity can be found using the Pythagorean theorem:
V_total = sqrt(V_horizontal^2 + V_vertical^2)
where:
V_total = magnitude of the total velocity
V_horizontal = speed of the plane (169 m/s)
V_vertical = vertical component of the velocity (calculated using the previous step)
4. To determine the horizontal distance traveled by the second package when thrown downward, we can again use the equation of motion:
distance = speed * time
where:
distance = horizontal distance traveled by the second package
speed = horizontal speed of the second package (which is the same as the speed of the plane, 169 m/s)
time = time for the package to reach the ground (calculated in the first question)
By following these steps, we can calculate the answers to all the questions.