How Does a Cowboy Time His Jump to Land on a Galloping Horse?

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Homework Help Overview

The problem involves a cowboy attempting to time his jump to land on a galloping horse. The horse is moving at a speed of 10 m/s, and the cowboy is positioned 3.0 m above the saddle. The discussion focuses on determining the horizontal distance required for a successful landing and the time the cowboy spends in the air.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to calculate the time it takes for the cowboy to fall 3.0 m and question the relevance of the horse's speed in this calculation. There is confusion regarding the initial vertical speed of the cowboy and how to apply the kinematic equations correctly.

Discussion Status

Some participants have provided guidance on treating vertical and horizontal motions independently. There is ongoing exploration of the correct application of kinematic equations, and one participant has suggested a time of 0.78 seconds for the fall, prompting further questions about its validity.

Contextual Notes

Participants are navigating common misconceptions related to projectile motion and the independence of vertical and horizontal components. The initial vertical speed of the cowboy is under discussion, and there is uncertainty about the implications of the calculated time on the jump's success.

soccergirl14
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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!
 
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soccergirl14 said:
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
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.)
 
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?
 
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?
 
soccergirl14 said:
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?

That's the idea.
If not, he missed the saddle :-p
 
haha, alright :)
thank you so much!
 

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