In the last second of free fall, a body covered 34th of its total path. Then the height from which the body is released will be  g=10 m/s2

If ‘n’ is the last second, then the distance travelled by a freely falling body in ‘n’ sec iss=12gt2 (∵u=0)=12gn2……..(1) and sn=gn−12……...(2) Given that sn=34s⇒gn−12=3412gn2∴(2n−1)=34n2⇒(2n−1)n2=34⇒n=2sThe height from which the body is released is s=12gn2                                                                           =12×10×(2)2=20m