# Sliding down a roof

1. Nov 1, 2007

### Vanessa23

1. The problem statement, all variables and given/known data
a ball started at a height of 12.5 meters and rolled to the edge of a roof which is at 7 meters. Using a^2+b^2=c^2 we know it rolled a distance of 8.14 meters on the roof because it rolled a horizontal distance of 6m and a vertical distance of 5.5m. Find the time it rolled down the roof. The roof makes a 40deg angle with the horizontal and coeff of friction is .388.

2. Relevant equations
Potential E=KE+Ffriction
mg*distance*sin(theta)=.5mv^2+(coeff of friction)*mgcos(theta)*distance down roof
acceleration=g*sin(theta)-g*(coeff. of friction)*cos(theta)
V=Vo+at
3. The attempt at a solution

8.14*g*sin(40)=.5v^2+.388*g*cos(40)*(8.14)
v=9.84m/s

a=g*sin40)-g*.388cos(40)
a=3.39m/s^2

9.84=3.39t
t=2.90 seconds

Last edited: Nov 1, 2007
2. Nov 2, 2007

### saket

What you have solved is for sliding down the roof whereas, the question has reiterated the term rolling. I think, question has been manipulated, in your own words!

The attempt is correct, method-wise, if it is sliding. So, have you written 'sliding' incorrectly?
If it is indeed 'rolling', one information is missing: information about the ball? Assuming a spherical ball (as most of them are), is it solid or hollow? Have you missed that?