# Vertical Drop and Speed

by waldvocm
Tags: speed, vertical
 P: 76 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.
 HW Helper PF Gold P: 3,440 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?
P: 11
 Quote by waldvocm 1. The problem statement, all variables and given/known data ...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?

 P: 76 Vertical Drop and Speed 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
 HW Helper PF Gold P: 3,440 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?
 P: 76 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.
 HW Helper PF Gold P: 3,440 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.
 P: 76 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
 P: 76 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
 HW Helper PF Gold P: 3,440 Yes, you can find the time this way and that would be the answer to part (b) incidentally.
 P: 76 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
 P: 76 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.
HW Helper
PF Gold
P: 3,440
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.
 Quote by waldvocm ... and the distance from the limb to the saddle is 3.84m.
Stop going around in circles.

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