(adsbygoogle = window.adsbygoogle || []).push({}); 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

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# Homework Help: Sliding down a roof

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