A plane flies at 20.0m/s ascending at 30 degrees above the horizontal. The plane drops a care package at a height that allows the package to land 750m away (horizontally).
1.) At what location above the ground was the package released?
2.) Find the velocity (magnitude and direction with respect to the horizontal) of the package at 50m above the ground.
3.) Sketch the horizontal and vertical component of velocity vector (separately) as the function of time
the horizontal speed is 17.32 m/s
So, the package falls for 750/17.32 = 43.30 seconds
So, you need h where
h + 10.0t - 4.9t^2 = 0
and t = 43.3
Now you can carry on with the other parts,
To solve these questions, we need to break down the given information and use relevant physics principles. Let's go step by step:
1.) At what location above the ground was the package released?
To determine the location above the ground where the package was released, we need to find the vertical distance traveled by the package from the release point to the landing point.
Since the plane is ascending at an angle of 30 degrees above the horizontal, we can use trigonometry to find the vertical distance. The vertical component of velocity can be found using the formula:
Vertical component of velocity = Initial velocity * sin(angle)
Here, the initial velocity is 20.0 m/s and the angle is 30 degrees.
Vertical component of velocity = 20.0 m/s * sin(30 degrees) = 10.0 m/s
The time taken for the package to travel horizontally is the same as for the plane. Therefore, we can also use the formula:
Vertical distance = Vertical component of velocity * time
We need to find time, which is equal to the horizontal distance (750 m) divided by the horizontal component of velocity (20.0 m/s).
Time = Horizontal distance / Horizontal component of velocity = 750 m / 20.0 m/s = 37.5 s
Now, substituting the value of time, we can find the vertical distance traveled by the package:
Vertical distance = 10.0 m/s * 37.5 s = 375 m
Therefore, the package was released at a location 375 meters above the ground.
2.) Find the velocity (magnitude and direction with respect to the horizontal) of the package at 50m above the ground.
To find the velocity (magnitude and direction) of the package at 50 meters above the ground, we need to consider the vertical and horizontal components of the velocity separately.
The magnitude of the velocity is the square root of the sum of the squares of the vertical and horizontal component velocities:
Magnitude of velocity = √(Vertical component of velocity^2 + Horizontal component of velocity^2)
The vertical component of velocity remains constant throughout the motion and is equal to 10.0 m/s.
The horizontal component of velocity can be found using the formula:
Horizontal component of velocity = Initial velocity * cos(angle)
Here, the initial velocity is 20.0 m/s and the angle is 30 degrees.
Horizontal component of velocity = 20.0 m/s * cos(30 degrees) = 17.32 m/s
Now, we can calculate the magnitude of the velocity:
Magnitude of velocity = √(10.0 m/s^2 + 17.32 m/s^2) = √(100.0 m^2/s^2 + 300.0 m^2/s^2) = √400.0 m^2/s^2 = 20.0 m/s
The direction with respect to the horizontal can be found using trigonometry:
Direction = arctan(Vertical component of velocity / Horizontal component of velocity)
Direction = arctan(10.0 m/s / 17.32 m/s) = arctan(0.577) = 29.7 degrees
Therefore, the velocity of the package is 20.0 m/s directed at an angle of 29.7 degrees above the horizontal.
3.) Sketch the horizontal and vertical component of velocity vector (separately) as the function of time.
To sketch the horizontal and vertical components of the velocity vector as the function of time, we need to understand how these components change over time.
The vertical component of velocity remains constant throughout the motion and is equal to 10.0 m/s, as calculated earlier. Therefore, the vertical component of the velocity remains a horizontal line on the graph.
The horizontal component of velocity also remains constant throughout the motion and is equal to 20.0 m/s, as given in the question. Therefore, the horizontal component of the velocity remains a horizontal line on the graph.
The vertical component graph will be a straight horizontal line at 10.0 m/s, and the horizontal component graph will be a straight horizontal line at 20.0 m/s.
Please note that the time scale should be based on the time taken for the package to travel from its release point to the desired height, as calculated earlier.
By sketching these two lines separately against the time axis, you will have the vertical and horizontal components of the velocity vector as functions of time.