Question

If a body travels half its total path in the last second of its fall from rest, find the time and height of its fall. Take g=10m/s²​

Answer 1

Correct Answer is: (2) 3.41 s, 57 m If the body falls a height h in time t, from 2nd equation of motion we have h = 1/2 gt2 ….(A) [u = 0 as body starts from rest] Now the distance fallen in (t – 1) s will be h = 1/2 g(t – 1)2 …(B) So from eq. (A) & (B) distance fallen in the last second h – h’ = (1/2) gt2 – (1/2) g (t – 1)2, h – h’ = (1/2) g (2t – 1) But according to given problem as (h – h’) = h/2 0.59 s is physically unacceptable as it gives the total time t taken by the body to reach ground is lesser than one sec while according to the given problem time of motion must be greater than 1 s. So t = 3.41 s & h = (1/2) × (9.8) × (3.41)2 = 57 mRead more on Sarthaks.com - https://www.sarthaks.com/571302/body-travels-half-total-path-last-second-fall-from-rest-time-height-fall-will-respectively