Finding the distance for one stone to pass the other

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SUMMARY

The problem involves two stones: the first stone is dropped from a cliff with an initial velocity of 0 m/s, while the second stone is thrown downward 1 second later with an initial velocity of 16 m/s. The acceleration due to gravity is -9.8 m/s² for both stones. The solution reveals that the second stone overtakes the first at a distance of 15.7 meters below the top of the cliff.

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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 = volt + 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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