Question

A body is released from rest to fall freely under gravity reaches the ground in 4 seconds find the height from which it was released

Answer 1

Add the total time of accelerated motion (2 secs) and remaining time (t secs) as uniform motion. The total distance covered in 125m. Of which in first 2 secs, the body covers -  s=12gt2 (at body starts from rest, u=0)  s=12∗9.8∗(2)2=19.6m  In the remaining t secs, it has to cover d=125−19.6=105.4m. The time t taken for this remaining journey is -  t=d/v, where v is the speed the body has reached by the end of 2 secs, which is v=gt (since initial speed u=0)  v=9.8∗2=19.6m/s  Now we can find time t -  t=d/v=105.4/19.6=5.37secs  The total time taken is 2+5.37=7.37 sec

Answer 2

Given data
g=a=9.8m/s²
d=s=?
t=4sec
u=0

by Newton's law of motion
s=ut+½at²
=0×4+1/2×9.8×4
=9.8×8
=78.4

hence, height = 78.4units

Hey mate! hope it helps you :-)