<p>So I’m self-studying Physics C: Mechanics and there’s a problem that I just can’t figure out. I’m either overthinking it or simplifying it too much… any help on it would be fantastic!!</p>
<p>Two stones are dropped from the edge of a 60 meter cliff, the second stone 1.6 seconds after the first. How far below the top of the cliff is the second stone when the seperation between the two stones is 36 meters?</p>
<p>By the way, if I remember correctly, the answer is 11 meters.</p>
<p>Ok, first let’s draw a picture</p>
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<p>Err, this is not going to work…</p>
<p>I assume you know how to draw the picture. </p>
<p>So, let d2 be the distance for the first falling rock and d1 be the distance for the second.</p>
<p>d2 = 1/2 g (t+1.6)^2 and d1 = 1/2 g t^2</p>
<p>so d2-d1=36 = 1/2 g ( (t+ 1.6)^2 -t^2)</p>
<p>then t = 1.45</p>
<p>Plugging back in for the second falling rock, 1/2 g *1.45^2 = around 10.5 (using calculator on the computer and used 10 not 9.8 for g)</p>