A block of ice starts sliding down from the top of an inclined roof of a along a line of the greatest slope. The inclination of the roof with the horizontal is `30^(@)`. The heights of the highest and lowerst points of the roof are 8.1 m and 5.6 m respectively.At what horizontal distance from the lowest point will the block hit the ground ? Neglect air friction .`[g=9.8m//s^(2)]`.

Explanation

Acceleration of the block along the greatest slope in equal to `a=g sin 30^(@)`
Distance travelled by the block along the greatest slope is equal to `S=((8.1-5.6))/(sin 30^(@))=5m`
If u be the speed of the block when it is just abou to leave the roof then
`u^(2)=0+2gsin 30^(@)xx5impliesu=7m//s`
If t be the time taken on hit the ground then

`5.6=u sin 30^(@)t+1/2"gt"^(2)=7/2t+1/2xx9.8t^(2)`
`implies7t^(2)+5t-8=0`
`t=(-5+-sqrt(25-(4)(7)xx(-8)))/(2xx7)`
`impliest=(-5+-15.78)/14s`, -ve value is to be rejected , i.e. `t=(-5+15.78)/14=0.77` sec
Horizontal distance travelled is equal to
`x=u cos 30^(@)t=(7sqrt(3))/2xx10.78/14mimpliesx=4.67m`