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.

Options

A. 4.33m
B. 14.33m
C. 8.998m
D. 0.668m

Explanation

The ice block sliding on roof AB travels a vertical distance
AC`=(8.1-5.6)m=2.5m`
If v is the velocity of block at B then we have the relation
`" " v^(2) = u^(2)+ 2as`
`u= 0 s=h_(1)=AC=2.5m`
and a `=g= 9.8m//s`
`:. V^(2)= 0+2 xx9.8xx2.5`
`implies v = 7m//s along BG`
The horizontal and vertical (downwards) components of velocity are given by

`v_(z) = v cos 30^(@) =7 xx sqrt3/2=3.5 sqrt3m//s`
`v-(z) vsin 30 ^(@) =7 xx 1//2= 3.5m//s`
The time taken by the block to reach the ground is determined by vertical motion by the relation
`S= UT + (1//2) at^(2)`
Here S`=h _(2) = 5.6, u=v _(y) = 3.5m//s `
`a = g = 9.8m//s`
`:> 5.6 = 3.5 t + (1//2) 9.8t^(2)`
`implies t=0.77s`
Horizontal distance
EF` = x_(2) = v _(20 t = 3.5 sqrt 3 xx 0.77 = 4.668m `
Also `x_(1) = ED = BC = AC cot 30^(@) = 2.5 sqrt3 =4.33`
` :.` Total distance
`x= x_(1) + x_(2) `
`=4.33+ 4.668`
`8.998m`