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I've been struggling with this for a while but despite my best efforts I have not been able to form suitable equations. Question is as follows:

A stone is dropped from a cliff $160 ~\text{ft}$ high and $\frac{1}{2}$ second later another stone is thrown vertically upwards from the base of the cliff at $64 ~\text{ft}~\text{sec}^{-1}$. Find the distance from the base of the cliff at which the two stones meet and their velocities at that moment.

I tried the following approach:

$$s_1=16t^2$$ $$s_2=64\left(t-\frac{1}{2}\right)-16\left(t-\frac{1}{2}\right)^2$$

where $$s_2=160-s_1$$

I don't think equations are correct but am not sure where I'm going wrong; the answer gives $60~\text{ft}$ for distance but I can't get this despite several attempts. I've done several questions where stones are released from the same position but not when they're in different positions. I don't know if this is a question in which relative motion can be used but I don't know how to approach the question if this is the case as velocities aren't constant. As always any advice would be greatly appreciated.

GR L
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  • I think the answer is incorrect or it is correct if both stones are released at same time. – GR L Nov 21 '21 at 18:10
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    As a good problem solving technique, make sure you clearly define the variables you're utilizing. In your work, $s_1$ is the distance the first stone travelled downward, $s_2$ is the distance the second stone travelled upward, and $t$ is the time since the first stone was dropped from the cliff. Am I correct? I'm getting an answer of $63.96$ feet. – Matthew H. Nov 21 '21 at 18:17
  • @MatthewPilling Yes that is correct and I get that answer also. Thank you for replying. – GR L Nov 21 '21 at 18:20

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