A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25m below the top. Both stones reach the bottom of  building simultaneously. The height of the building is

Time of release of  2nd stone is when 1st stone falls for 5m.∴5=12gt2⇒t=2×510=1sLet say in time t 1st stone from top reaches to ground. So, 2nd stone will reach ground in (t-1).∴disp of 1st stone = disp of 2nd stone + 25⇒12gt2=12g(t−1)2+25⇒5t2=5t2+1−2t+25⇒10t=30⇒t=3sec For 1st stone, h=12gt2⇒h==12×10×9=45m