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).

An object falling freely covers half of its total distance in last sec

1.	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).
Answer» Solution :Suppose, total time is T and total height is h, In `h = ut + 1/2 G g ^(2) , u = 0` `h = 1/2g 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`