Calculating the Intersection of Two Stones Dropped from a Cliff

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To find the intersection point of the two stones, first establish the equations of motion for both stones. The stone dropped from the cliff has an initial velocity of 0 and accelerates downward due to gravity, while the stone projected upwards has an initial velocity of 96 ft/s and also experiences gravitational acceleration. Set the distance traveled by the upward stone equal to 256 feet minus the distance traveled by the downward stone to find the meeting point. By substituting the equations of motion into this relationship, solve for the time at which they meet. This will allow you to calculate the height at which the two stones intersect.
John O' Meara
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A stone is droped from the top of a cliff of height 256 feet, and, at the same instant, another stone is projected vertically upwards from the ground with a speed of 96 ft/s. Find where they will meet?
I have the following:
v1=u1+at, v2=u2+at, s1=u1t+1/2at^2, s2=u2+1/2at^2,
I also have: v1^2=u1^2+2as1 and V2^2=u2^2+2as2
The question is what do I do next, please?
 
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You have the inital velocity and acceleration of both. You also know that the distance the stone thrown upwards from the ground has traveled when they meet is 256-s where s is the distance traveled by the dropped stone.
 
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