A block of ice is sliding down the sloping roof of a house and the angle of inclination of the roof with the horizontal is `30^@` . The maximum and minimum heights of the roof from the ground are 8.1 m and 5.6 m. How far from the starting point, measured horizontally , does the block land ? [ignore friction]

Explanation

Let the highest point of the roof be A and the lowest point be P as shown in Fig.2.69.
`therefore AC=8.1m, PD =5.6m`
`therefore AB=AC-BC=AC-PD=8.1-5.6=2.5m`
`AP=(AB)/(sin 30^@)=(2.5)/(1/2)=5m`

Let the velocity of the block at P be v .
Considering the motion of the block from A to P,
`v^2=2g sin 30^2xxAP =2xx9.8xx1/2xx5 =49 or, v=7m*s^(-1)`.
The horizontal and vertical components of the velocity at P are `v cos 30^@ =(7sqrt(3))/(2) m*s^(-1) and v sin 30^@ =7/2 m*s^(-1)` respectively.
Let the total taken by the block to come from P to E be t. Considering the vertical motion of the block ,
`5.6=7/2t+1/2xx9.8xxt^2 or, 7t^2 +5t-8=0`
`therefore t=(-5pmsqrt(25+224))/(14)=0.77s` [taking the positive value of t]
Now, `DE=v cos30^@ xxt=(7sqrt(3))/(2)xx0.77=2.7sqrt(3) m`
`therefore CE=CD+DE=2.5sqrt(3)+2.7sqrt(3)`
[`tan 30^@ =(AB)/(PB) or, 2.5 sqrt(3)m and , DC =PB =2.5 sqrt(3)m]`
=`5.2 sqrt(3)=9m`.