An object falling freely covers half of its total distance in last second, then find total height and total times g = 9.8 `ms^(-2).`

Explanation

Suppose, total time is T and total height is h,
In `h = ut + 1/2 g g ^(2) , u = 0`
`h = 1/2 g t ^(2) " "…(1)`
Now `t = (T-1) and h = h/2`
`h /2 = 1/2 g (T -1) ^(2)`
`therefore h = g (T - 1) ^(2) " "…(2)`
By comparing equation (1) and (2).
`therefore 1/2 g t ^(2) = g (T ^(2) - 2T +1)`
`therefore T ^(2) = 2 T ^(2) - 4 T + 2`
`therefore T ^(2) -4T + 2=0`
`a = 1, b =- 4, c =2`
`Delta = b ^(2) - 4ac`
`=16-4 xx 1 xx 2 `
`=16 -8 =8`
`therefore sqrtDelta = 2 sqrt2`
`therefore` Solution `T = (-b pm sqrtDelta )/( 2a) = (4 pm 2 sqrt2)/(2)`
`T = 2 pm sqrt2`
`therefore T = 2 + 1.4.4 = 3.414s or 2-1414 = 0.586 s`
But `T=0.586` s is not possible
`therefore T = 3.414 s `
Now from equation `h = 1/2 g t ^(2) `
`h = 1/2 xx 9.8 xx (3.414) ^(2)`
`therefore h = 57.11 m`