How can the ranch hand successfully jump onto the galloping horse?

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Homework Help Overview

The discussion revolves around a physics problem involving kinematics, specifically the scenario of a ranch hand attempting to drop onto a galloping horse. The horse is moving at a speed of 9.28 m/s, and the vertical distance from the tree limb to the saddle is 3.84 m. Participants are exploring how to calculate the necessary horizontal distance for the jump.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the time it takes for the ranch hand to drop to the saddle and question the method used to calculate this time. There are attempts to apply kinematic equations to find the relationship between distance, time, and velocity. Some participants express confusion about the assumptions regarding the problem setup and the need for additional information.

Discussion Status

The discussion is ongoing, with participants providing guidance on the use of kinematic equations and clarifying the problem's parameters. There is recognition of the need to find the time of the drop and the horizontal distance, but no consensus has been reached on the best approach to solve the problem.

Contextual Notes

Some participants note the difficulty of self-teaching the material without direct instruction, highlighting the challenges faced in an online learning environment. There is also mention of confusion regarding the interpretation of the problem's details, particularly the distance between the limb and the saddle.

waldvocm
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Homework Statement


A ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree at the speed of 9.28 m/s and the distance from the limb to the saddle is 3.84m.

a) What must be the horizontal distance (m) between the saddle and the limb when the ranch hand makes his move?


Homework Equations


It will take the person .39 s to drop from the tree limb to the saddle. So when the horse it .39 seconds away the person should jump. Should I use an equation position as a function of time?


The Attempt at a Solution



3.88m I divided .39 by 9.28 = .042 then added that to the original distance away from the saddle leaving me with 3.88m as a final answer.
 
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How did you figure it will take the person 0.39 s to drop down? Dividing 3.84 m by 9.8 m/s2 is not the way to go. What kinematics equation is applicable here?
 
waldvocm said:

Homework Statement


...the distance from the limb to the saddle is 3.84m.

Without a picture one can only assume what it means. Is it saying when the saddle is vertically underneath the tree limb that the distance between the limb and the saddle is 3.84m?
 
Yes, that is what I am assuming.

Would I use the kinematics equation for position as a function of velocity and time.

Xf = Xi + 1/2(Vxi + Vxf)t
 
You don't know two things in that equation, time and final velocity. What other kinematics equation relates displacement and time and also involves the acceleration which is known?
 
Vxf^2 = Vxi^2 + 2ax(d)

Vxf^2 = 0 + 2(9.28)(3.84)

I thought this was right, but shouldn't I be solving for the displacement.
 
Ok, this gives you his speed when he hits the saddle. How are you going to find the time it takes him to drop down?

You should not be solving for the displacement because you that it is 3.84 m.
 
Do I need to know the time it takes him to drop down?

This is the questions - What must be the horizontal distance (m) between the saddle and the limb when the ranch hand makes his move?

Part b asks how long he is in the air
 
to find the time I could solve the previous equation for the final V and use this equation -

Vf = Vi + a*t

to find the time he is in the air
 
  • #10
Yes, you can find the time this way and that would be the answer to part (b) incidentally.
 
  • #11
I am still confusec as to how to find the distance between the limb and the saddle when he makes his move.

Can I use the variables that I just found Vf and t to solve for the displacement?

Xf = Xi + 1/2(Vxi + Vxf)t
 
  • #12
Thank you for your time and patients. I am taking this class online and I am finding the lack of lecture and one on one help to be very challenging. It is difficult material to teach yourself out of a book.
 
  • #13
Yes, you can use that equation to find the time.
I am still confusec as to how to find the distance between the limb and the saddle when he makes his move.

What's so confusing? The problem gives you the distance between the limb and the saddle, you don't need to find it.
waldvocm said:
... and the distance from the limb to the saddle is 3.84m.

Stop going around in circles.
 

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