Two stone fall down a shaft, the second one beginning its fall 1 sec after the first. Find the second stone’s motion in relation to that of the first. Ignore air-resistance.
Text Solution
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Sol. Both stones move relative to the earth with the same constant and uniform acceleration g. Clearly one stone move uniformly in relation to the other, and the constant speed of the first stone acquires in 1 sec., i.e. in the period that elapses between the two moments at which the stones start falling.
It is not difficult to carry out the necessary calculation.
The distance traveled by the first stone is found from the equation
s 1 = 
The distance traveled by the second stone from the equation s 2 = 
.
The distance between the two stones increases with the lapse of time according to the formula
s 1 – s 2 = gt – 
,
i.e., the first stone moves uniformly in relation to the second stone with a velocity numerically equal to g.
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