What is the Distance Between Two Stones Dropped from a Well in One Second?

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The problem involves calculating the distance between two stones dropped from a well, with the first stone dropped at time t=0 and the second at t=1 second. The equations of motion are applied, specifically using the formula for distance under constant acceleration due to gravity. The initial calculations yield an incorrect answer, prompting a review of the gravitational constant used and the need for unit conversion from meters to feet. Ultimately, the correct distance between the stones after two seconds is determined to be 48.3 feet.
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Homework Statement


A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later.
The well is 80ft deep


Homework Equations


V=V0+at
X =X0+V0t+.5at^2


The Attempt at a Solution


nce apart= distance first-distance2nd
= 1/2 g t^2 - 1/2 g(t-1)^2
so, at t=2
= 1/2 g 4 - 1/2 g 1
this is not giving me the correct anwser. Correct anwser is 48.3ft
 
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myoplex11 said:
dnce apart= distance first-distance2nd
= 1/2 g t^2 - 1/2 g(t-1)^2
so, at t=2
= 1/2 g 4 - 1/2 g 1
this is not giving me the correct anwser. Correct anwser is 48.3ft

Hi!

Looks ok to me. What figure are you using for g? :confused:
 
Your answer is correct. Don't forget to convert meters to feet.
 
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