Find the initial velocity of the package relative to the ground

In summary, a helicopter is flying at a constant speed and altitude, and a package is ejected from it with an initial velocity in the opposite direction. To find the initial velocity relative to the ground, we use the double-subscript convention to add velocities. To find the horizontal distance between the helicopter and the package, we cannot use the Pythagorean theorem and must consider kinematic equations for projectile motion. By setting values into x and y components, we can find the initial vertical velocity and the time it takes for the package to drop. These values can then be used to answer the remaining questions about the package's velocity and angle before impact.
  • #1
Kpgabriel
36
0

Homework Statement


A helicopter is flying in a straight line over a level field at a constant speed of 6.8 m/s and at a constant altitude of 9.4 m. A package is ejected horizontally from the helicopter with an initial velocity of 12.0 m/s relative to the helicopter, and in a direction opposite the helicopter's motion. Find the initial velocity of the package relative to the ground.

What is the horizontal distance between the helicopter and the package at the instant the package strikes the ground?

What angle does the velocity vector of the package make with the ground at the instant before impact, as seen from the ground?

Homework Equations


Initial Velocity:
Vhg = 6.8 m/s
Vph = -12.0 m/s
Vpg = -5.20 m/s
Y = 9.8 m

The Attempt at a Solution


I was thinking that I could use a kinematic equation to solve for this question but I think it is much simpler than that.
 
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  • #2
Did you have a question? You haven't given us anything we can examine or comment on in terms of a solution attempt. Can you show in detail what you've tried?
 
  • #3
It is. How did you get Vpg which is what you are looking for?
 
  • #4
kuruman said:
It is. How did you get Vpg which is what you are looking for?
To find Vpg I set it equal to Vhg + Vph. I am not sure why I did that. I only know that the package is going opposite of the helicopter so its velocity is negative.
Vpg = Vhg +Vph
 
  • #5
Are you familiar with the double-subscript convention for adding velocities? Most textbooks have it. Check yours.
 
  • #6
gneill said:
Did you have a question? You haven't given us anything we can examine or comment on in terms of a solution attempt. Can you show in detail what you've tried?
To find the horizontal distance I tried Pythagorean theorem: x^2 + 9.8^ = -12.0 ^2 and I got 6.8
 
  • #7
kuruman said:
Are you familiar with the double-subscript convention for adding velocities? Most textbooks have it. Check yours.
Would that help with finding the horizontal displacement?
 
  • #8
kuruman said:
Are you familiar with the double-subscript convention for adding velocities? Most textbooks have it. Check yours.
would it be Xhp = Xhg + Xgp?
 
  • #9
Kpgabriel said:
Would that help with finding the horizontal displacement?
It would because it gives you the initial velocity relative to the ground. However, it looks like you found that and your question is about the horizontal distance
Kpgabriel said:
To find the horizontal distance I tried Pythagorean theorem: x^2 + 9.8^ = -12.0 ^2 and I got 6.8
To answer that, you cannot use the Pythagorean theorem because the path is a parabola not a straight line. You need to consider the kinematic equations as modified for projectile motion.
 
  • #10
kuruman said:
It would because it gives you the initial velocity relative to the ground. However, it looks like you found that and your question is about the horizontal distance

To answer that, you cannot use the Pythagorean theorem because the path is a parabola not a straight line. You need to consider the kinematic equations as modified for projectile motion.
If that's the case usually I like to set my values into x and y components:
Y:
(Package)
Y = 9.8m
Vi = -5.20 m/s
Vf = -12.83 m/s solved by kinematic equation V^2 = Vi^2 + 2s(Y - Yi)
g= -9.8 m/s^2
t = 0.78s ~ V = Vi + at
X:
V = 6.8 m/s helicopter
 
  • #11
Kpgabriel said:
If that's the case usually I like to set my values into x and y components:
Y:
(Package)
Y = 9.8m
Vi = -5.20 m/s
Vf = -12.83 m/s solved by kinematic equation V^2 = Vi^2 + 2s(Y - Yi)
g= -9.8 m/s^2
t = 0.78s ~ V = Vi + at
X:
V = 6.8 m/s helicopter
Ok so from my time I am multiplying it to the helicopters velocity because it is a constant and does not change, which means acceleration is 0, and end up getting 7.58 m for x but it is wrong.
 
  • #12
Kpgabriel said:
Vi = -5.20 m/s
If the helicopter is traveling at constant altitude, what is the package's initial vertical velocity?
 
  • #13
kuruman said:
If the helicopter is traveling at constant altitude, what is the package's initial vertical velocity?
Is it 0 m/s? because it is not moving yet.
 
  • #14
Correct. So in the vertical direction you have a package that starts with zero velocity and drops 9.8 m. Can you find how much time it takes to do that?
 
  • #15
kuruman said:
Correct. So in the vertical direction you have a package that starts with zero velocity and drops 9.8 m. Can you find how much time it takes to do that?
yes it takes 1.41s from V = Vi +at. V is now -13.86 m/s because the initial Velocity was 0 m/s
 
  • #16
Which V is this? If it is a component, it needs a subscript x or y. Is it a component?
 
  • #17
kuruman said:
Which V is this? If it is a component, it needs a subscript x or y. Is it a component?
Its the final velocity of the package so Vy
 
  • #18
Correct. Add the subscript so that you don't get confused.
Now you have enough information to answer the remaining two questions. Do you see how?
 
  • #19
kuruman said:
Correct. Add the subscript so that you don't get confused.
Now you have enough information to answer the remaining two questions. Do you see how?
Yes. Thank you for your help. Sorry for the late response I went to a research meeting.
 

Related to Find the initial velocity of the package relative to the ground

1. What is the definition of initial velocity?

The initial velocity of an object is the speed and direction at which it is moving at the beginning of its motion.

2. How is initial velocity different from average velocity?

Initial velocity refers to the speed and direction at the start of an object's motion, while average velocity is the average speed and direction over a specific time interval.

3. How is the initial velocity of a package relative to the ground calculated?

The initial velocity of a package relative to the ground can be calculated by using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time elapsed.

4. Can initial velocity be negative?

Yes, initial velocity can be negative if the object is moving in the opposite direction of the chosen positive direction. For example, if the positive direction is towards the east, a package moving towards the west would have a negative initial velocity.

5. How can initial velocity affect the trajectory and distance traveled by a package?

The initial velocity of a package can greatly impact its trajectory and distance traveled. A higher initial velocity will result in a longer distance traveled and a flatter trajectory, while a lower initial velocity will result in a shorter distance traveled and a steeper trajectory.

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