# A bushranger hanging out of a hotel window

1. Mar 23, 2013

### PG222

1. The problem statement, all variables and given/known data
A bushranger hanging out of a hotel window wishes to drop vertically onto a horse galloping under the window. The constant speed of the horse is 14.0 m/s, and the distance from the window to the level of the saddle is 2.50 m.
(a) What must be the horizontal distance between the saddle and window sill when the bushranger makes his move?

(b) For what time interval is he in the air?

2. Relevant equations
I'm not too sure of what equations I was meant to use but I tried this:
X=-2.5
X=0.5(-9.8)t^2
D=Ut
3. The attempt at a solution
Using the above equations I worked t out to be 0.714 seconds
And for the horizontal distance I ended up with 10 meters

Last edited: Mar 24, 2013
2. Mar 24, 2013

### Simon Bridge

Welcome to PF;
Was there a question in all that?

some things stand out:
1. you have not provided the reasoning for your working
2. you have not done anything for part (a)
3. would "distance" be the magnitude of the displacement here?

Kinematic equations would be appropriate here - especially if you want to neglect air resistance.