A block of ice starts sliding down from the top of the inclined roof of a house (angle of inclination of roof=`30^(@)` with the horizontal) along aline of maximum slope. The highest and lowest points of the roof are at heights of 8.1m and 5.6 m respectively from the ground. At what horizontal distance from the starting point will the block hit the ground? Neglect friction.

Explanation

When the ice block silides down from A to B, it falls vertically through AC=2.5m
If v is the velocity with which it reaches B, then `v^(2)=u^(2)+2as`, where u=0,a=`g sin 30^(@)` and s=AB.
`thereforev^(2)=2gxx(AC)/(AB)xxAB(sin 30^(@)=(AC)/(AB))`
`=2xx9.8xx2.5=49`
`therefore v=7ms^(-1)` along AB, Resolve this velocity v into:
(i) a horizontal component
`=v cos 30^(@)=7xx(sqrt(3)//2)=3.5xxsqrt(3)ms^(-1)`
Vertical motion:
Let t be the time taken by the block to reach the ground.
Apply `s=ut+(1)/(2)g t^(2)`
Given `s=5.6m,u=3.5ms^(-1),g=9.8ms^(-2)`
`therefore 5.6=3.5t+(1)/(2)xx9.8t^(2)` or `4.9t^(2)+3.5t-5.6=0`
Dividing this equation by 0.7, we get `7t^(2)+5t-8=0`
`therefore t=(-5+0sqrt((5)^(2)+(4xx7xx8)))/(2xx7)`
`therefore t=(-5+-15.78)/(14)`
Rejecting the -ve value,
`=(-5+15.78)/(14)=(10.78)/(14)=0.77s`
Horizontal Motion:
Horizontal distance travelled EF
=Horizontal velocity xx time
`=3.5xxsqrt(3)xx0.77=4.668m`
`therefore` Total horizontal distance =DF=DE+EF
`=4.33+4.668=8.998m`
`(DE=BC=AC cot 30^(@)=2.5xxsqrt(3)=4.33m)`