A body travels half of its total path in the last second of its free fall from rest. The duration of fall is nearly?
Answers
Answered by
6
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
HOPE SO IT WILL HELP U....
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....
Anshikapotter:
Is this a formula?
Yes formula used is h=ut+1/2gt^2 where particle is at rest so u=0
Oh
Similar questions