PLEASE HELP! this is do tomorrow :( I have no clue how to do any of it.
A plane is flying horizontally with speed
193 m/s at a height 2010 m above the ground, when a package is dropped from the plane.The acceleration of gravity is 9.8 m/s^2.
1.What is the horizontal distance from the release point to the impact point?
Answer in units of m.
2.A second package is thrown downward from the plane with a vertical speed v1 = 58 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.
3.What horizontal distance is traveled by this package?
Answer in units of m
See previous post: Wed, 9-9-15, 5:19 PM.
To solve these problems, we can use the equations of motion to analyze the motion of the package. Let's break down each question step by step:
1. To find the horizontal distance from the release point to the impact point, we need to determine the time it takes for the package to reach the ground. We can use the equation:
h = 1/2 * g * t^2
where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
In this case, the height is given as 2010 m. Rearranging the equation to solve for t, we get:
t = sqrt(2h / g)
Substituting the given values, we have:
t = sqrt(2 * 2010 / 9.8) ≈ 20.1 s
Now, we can calculate the horizontal distance using the formula:
d = v * t
where d is the horizontal distance and v is the horizontal velocity (193 m/s).
Substituting the values, we get:
d = 193 * 20.1 ≈ 3879.3 m
Therefore, the horizontal distance from the release point to the impact point is approximately 3879.3 m.
2. To find the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground, we need to calculate the resultant velocity.
The horizontal velocity (v_x) remains constant at 193 m/s because there is no horizontal acceleration. The vertical velocity (v_y) is given as -58 m/s (downward direction).
The resultant velocity (v) can be found using the Pythagorean theorem:
v = sqrt(v_x^2 + v_y^2)
Substituting the given values, we have:
v = sqrt((193)^2 + (-58)^2) ≈ 203.9 m/s
Therefore, the magnitude of the total velocity of the package at the moment it is thrown, as seen by an observer on the ground, is approximately 203.9 m/s.
3. To find the horizontal distance traveled by the package, we need to determine the time it takes for the package to reach the ground. We already calculated this in question 1, which is approximately 20.1 s.
Now, we can calculate the horizontal distance using the formula:
d = v_x * t
where d is the horizontal distance, v_x is the horizontal velocity (193 m/s), and t is the time.
Substituting the values, we get:
d = 193 * 20.1 ≈ 3879.3 m
Therefore, the horizontal distance traveled by the package is approximately 3879.3 m, which is the same as the answer to question 1.
I hope this helps! If you have any further questions, feel free to ask.