A body released from certain height above the ground describes 716 of the total height in the last second of its fall. Then it is falling from a height equal to g=9.8 m/s2

Let ‘n’ be the last second of its fall the total height covered after ‘n’ sec ish=ut+12gt2⇒h=0+12gn2=12gn2………..(1)The distance travelled in the last second of its fall is  sn=u+gn−12⇒716h=0+g(2n−1)2⇒71612gn2=g(2n−1)2⇒2n−1n2=716On solving  n=4 s∴ The time taken by the body  t=4 s∴ The height  h=12gn2                                       =12×9.8×(4)2=9.8×8=78.4m