Asked by Lynn
A ball is dropped from a building and falls with an acceleration of magnitude 10m/s^2. The distance between floors in the building is constant. The ball takes 0.5s to fall from the 14th to the 13th floor and 0.3s to fall from the 13th floor to the 12th. What is the distance between floors?
No additional physics/forces need to be taken into consideration than what is described in the question.
No additional physics/forces need to be taken into consideration than what is described in the question.
Answers
Answered by
bobpursley
the time to fall to the 14th floor is
h=1/2 g (t^2)
the time tofall tothe 13th floor is
h+d=1/2 g (t+.5)^2
the time to fall to the 12th floor is
h+2d=1/2 g (t+.8)^2
subtract the first equation from the second..
d=1/2g(t+.25) check that.
now subtract the first equation from the third
2d=1/2 (1.6t+.64)
There are a number of ways to solve these two for d, t. The easiest I think it to solve for t in the first, then put that into the second and solve for d.
check my thinking
h=1/2 g (t^2)
the time tofall tothe 13th floor is
h+d=1/2 g (t+.5)^2
the time to fall to the 12th floor is
h+2d=1/2 g (t+.8)^2
subtract the first equation from the second..
d=1/2g(t+.25) check that.
now subtract the first equation from the third
2d=1/2 (1.6t+.64)
There are a number of ways to solve these two for d, t. The easiest I think it to solve for t in the first, then put that into the second and solve for d.
check my thinking
Answered by
Lynn
I'm sorry if I sound a tad confused, but what does g represent? Is it the acceleration (10 m/s^2)?
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