A particle is droped under gravety from rest a height h and it travels a distance (9h)/(25)in the last second, the height h is (take g=9.8ms^(-2))

A particle is droped under gravety from rest a height h and it travels

1.	A particle is droped under gravety from rest a height h and it travels a distance (9h)/(25)in the last second, the height h is (take g=9.8ms^(-2))
Answer» Solution :`(h)_(nth)=u+((2n+1))/2g"(or)"(9H)/(25)=0+((2n-1))/2g` `h=0+1/2gxx(n)^(2)` (1)//(2), we get `9/(25)=((2n-1))/n^(2)` `9n^(2)=50n-25` `9n^(2)-50n+25=0` After SOLVING `n=5s, 5/9s` `H=122.5m1.5m`.