# Vertical Drop and Speed

1. Oct 1, 2009

### waldvocm

1. The problem statement, all variables and given/known data
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?

2. Relevant 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?

3. 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.

2. Oct 1, 2009

### kuruman

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?

3. Oct 1, 2009

### librab103

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?

4. Oct 1, 2009

### waldvocm

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

5. Oct 1, 2009

### kuruman

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?

6. Oct 1, 2009

### waldvocm

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.

7. Oct 1, 2009

### kuruman

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.

8. Oct 1, 2009

### waldvocm

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

9. Oct 1, 2009

### waldvocm

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. Oct 1, 2009

### kuruman

Yes, you can find the time this way and that would be the answer to part (b) incidentally.

11. Oct 1, 2009

### waldvocm

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. Oct 1, 2009

### waldvocm

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. Oct 1, 2009

### kuruman

Yes, you can use that equation to find the time.
What's so confusing? The problem gives you the distance between the limb and the saddle, you don't need to find it.
Stop going around in circles.