Finding the distance for one stone to pass the other

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To determine the distance at which the second stone overtakes the first, the problem involves calculating the distance fallen by both stones under the influence of gravity. The first stone, dropped from rest, falls with an acceleration of -9.8 m/s², while the second stone, thrown downward with an initial speed of 16 m/s, also accelerates at the same rate. After 1 second, the first stone has fallen a distance of 4.9 meters, while the second stone begins its descent. By setting the equations of motion for both stones equal, the solution reveals that the second stone overtakes the first at a distance of 15.7 meters below the cliff top. This calculation confirms the expected answer.
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Homework Statement


A stone is dropped from the top of a tall cliff(with zero initial speed), and 1.00s later a second stone is thrown vertically downward w/ a speed of 16 m/s. How far below the top cliff will the second stone overtake the first

Homework Equations


distance = vot + 1/2a(tsquared)

The Attempt at a Solution


I attempted this so many different ways but I'm just not getting the answer. I know the vo of the first stone is zero and the acceleration is -9.8 m/s squared. I think that the second stones initial velocity is 16 m/s.
 
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answer should be 15.7 m by the way
 
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