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 is 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 just about to leave the roof then
`u^(2) = 0+ 2 g sin 30^(@) 30^(@) xx 5rArr u = 7 m//s`
If the be the time taken to hit the ground then
`5.6 = u sin 30^(@) t + (1)/(2) gt^(2) = (7)/(2)t + (1)/(2) xx 9.8 t^(2)`
`rArr 7t^(2) + 5t - 8 = 0`
` t = (-5+pm sqrt(25-(4)(7)xx(8)))/(2xx7)`
`rArr t = (-5 pm 15.78)/(14) ` , -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 = (7 sqrt(3))/(2) xx (10.78)/(14) m rArr x = 4.67 m `