SOLUTION: A stone is dropped from the top of a cliff 100m high, and 2 seconds later a second stone is thrown vertically downward with a velocity of 30m/s. How far below the top of the cliff

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Question 1087346: A stone is dropped from the top of a cliff 100m high, and 2 seconds later a second stone is thrown vertically downward with a velocity of 30m/s. How far below the top of the cliff will the second stone overtake the 1st stone?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Start measuring time when the first stone is dropped from the cliff, it falls under gravity
d%5B1%5D=100-%289.8%2F2%29t%5E2
d%5B1%5D=100-4.9t%5E2
The second stone starts at t=2 with an initial velocity of 30, also falling under gravity,.
d%5B2%5D=100-30%28t-2%29-%289.8%2F2%29%28t-2%29%5E2
d%5B2%5D=100-30%28t-2%29-4.9%28t-2%29%5E2
d%5B2%5D=100-30t%2B60-4.9t%5E2%2B19.6t-19.6
d%5B2%5D=140.4-10.4t-4.9t%5E2
So then find t when d%5B1%5D=d%5B2%5D,
100-4.9t%5E2=140.4-10.4t-4.9t%5E2
-10.4t=-40.4
t=3.885s
Now that you have the time, plug it into either distance formula and calculate the height from the ground in meters.