A body travels half of its total path in the last second of its free fall from rest . the duration of fall is nearly
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Answered by
147
suppose the body travelled total t second and the height is h then we have
1/2gt^2 = h
and
1/2 g(t-1) ^ 2=h/2=1/4 gt^2
=>g(t^2 + 1 - 2t) = 1/2gt^2
=>1/ 2gt ^2- 2gt+g=0
=> 3.41
As the body travelled more than 1 s so answer is 3.41s
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1/2gt^2 = h
and
1/2 g(t-1) ^ 2=h/2=1/4 gt^2
=>g(t^2 + 1 - 2t) = 1/2gt^2
=>1/ 2gt ^2- 2gt+g=0
=> 3.41
As the body travelled more than 1 s so answer is 3.41s
HOPE SO IT WILL HELP U....
IF IT HELPS MARK AS BRAINIEST......
Answered by
14
Answer:
answer of this question is 3.41sec,
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