Find the initial velocity of the package relative to the ground

[Kpgabriel]

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.

 

[gneill]

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?

 

[kuruman]

It is. How did you get Vpg which is what you are looking for?

 

[Kpgabriel]

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

 

[kuruman]

Are you familiar with the double-subscript convention for adding velocities? Most textbooks have it. Check yours.

 

[Kpgabriel]

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

 

[Kpgabriel]

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?

 

[Kpgabriel]

Are you familiar with the double-subscript convention for adding velocities? Most textbooks have it. Check yours.

would it be Xhp = Xhg + Xgp?

 

[kuruman]

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

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.

 

[Kpgabriel]

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

 

[Kpgabriel]

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.

 

[kuruman]

Vi = -5.20 m/s

If the helicopter is traveling at constant altitude, what is the package's initial vertical velocity?

 

[Kpgabriel]

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.

 

[kuruman]

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?

 

[Kpgabriel]

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

 

[kuruman]

Which V is this? If it is a component, it needs a subscript x or y. Is it a component?

 

[Kpgabriel]

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

 

[kuruman]

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?

 

[Kpgabriel]

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.