A body falls freely from a height of 125 m. after 2sec gravity ceases to act.find time taken to reach the ground
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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..
Hope this helps u!!
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..
Hope this helps u!!
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