Register to reply

Vertical Drop and Speed

by waldvocm
Tags: speed, vertical
Share this thread:
waldvocm
#1
Oct1-09, 12:09 PM
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.
Phys.Org News Partner Science news on Phys.org
FIXD tells car drivers via smartphone what is wrong
Team pioneers strategy for creating new materials
Team defines new biodiversity metric
kuruman
#2
Oct1-09, 12:35 PM
HW Helper
PF Gold
kuruman's Avatar
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?
librab103
#3
Oct1-09, 12:59 PM
P: 11
Quote Quote by waldvocm View Post
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?

waldvocm
#4
Oct1-09, 02:04 PM
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
kuruman
#5
Oct1-09, 02:13 PM
HW Helper
PF Gold
kuruman's Avatar
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?
waldvocm
#6
Oct1-09, 02:41 PM
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.
kuruman
#7
Oct1-09, 02:45 PM
HW Helper
PF Gold
kuruman's Avatar
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.
waldvocm
#8
Oct1-09, 02:59 PM
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
waldvocm
#9
Oct1-09, 03:02 PM
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
kuruman
#10
Oct1-09, 03:11 PM
HW Helper
PF Gold
kuruman's Avatar
P: 3,440
Yes, you can find the time this way and that would be the answer to part (b) incidentally.
waldvocm
#11
Oct1-09, 03:21 PM
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
waldvocm
#12
Oct1-09, 03:23 PM
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.
kuruman
#13
Oct1-09, 03:27 PM
HW Helper
PF Gold
kuruman's Avatar
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 Quote by waldvocm View Post
... and the distance from the limb to the saddle is 3.84m.
Stop going around in circles.


Register to reply

Related Discussions
To find the critical speed of a vertical shaft Mechanical Engineering 44
Vertical speed Introductory Physics Homework 2
Bernoulli's Pressure Drop Segregated from Friction Pressure Drop General Physics 7
Vertical SHM Introductory Physics Homework 28