# Kinematics Problem on cowboy

1. Mar 2, 2009

### soccergirl14

3. A daring cowboy sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The speed of the horse is 10. m/s, and the distance from the limb to the saddle is 3.0 m.
a. What must be the horizontal distance between the saddle and the limb when the cowboy makes his move?
b. How long is the cowboy in the air?

Okay, well I am really having troubles with this problem, I am not really sure where to start or what equations would be appropriate.

Here is my attempt:
I tried to find the time it takes to fall the 3.0m first, it didn't work too well.
d=.5at^2+Vit
3.0m= (-4.9m/s^2)t^2+(10m/s)t
now I am just confused, I don't even think I started the problem well. how do I solve for the t? Ahhh, please help!

2. Mar 2, 2009

### Staff: Mentor

When trying to figure out the time it takes for the cowboy to fall, the speed of the horse is irrelevant. And when you are computing the time it takes to fall a given distance, Vi is the initial vertical speed (of the cowboy). What's his initial speed?

(Also, since "down" is negative, the final position is -3.0 m.)

3. Mar 2, 2009

### CompuChip

Welcome to PF.
Don't worry, you are falling in a common trap :)
The trick to this kind of problems is, that you treat the vertical and the horizontal motions independently.
d = 0.5 a t^2 + Vi t
is a correct formula, but you are mixing things up after that.
What is the initial vertical velocity of the cowboy?

4. Mar 2, 2009

### soccergirl14

Okay, that makes a lot of sense, thank you!

Just another quick question, for time I got .78s, would that be the answer for B aswell? Since the cowboy is dropping 3.0m from the tree and that takes .78s, is that the time he is in the air?

5. Mar 2, 2009

### CompuChip

That's the idea.
If not, he missed the saddle :tongue:

6. Mar 2, 2009

### soccergirl14

haha, alright :)
thank you so much!