
#1
Feb2813, 12:12 AM

P: 12

1. The problem statement, all variables and given/known data
One stone is dropped from the the top of a tall cliff, and a second stone with the same mass is thrown vertically from the same cliff with a velocity of 10.0 m/s [down], 0.50seconds after the first. Calculate the distance below the top of the cliff at which the second stone overtakes the first? 2. Relevant equations v=0 d= 1/2(a)(t)^2 v=10 d=10(t)1/2(a)(t)^2 3. The attempt at a solution we have two unknowns, how can we solve it?!?! 



#2
Feb2813, 12:22 AM

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#3
Feb2813, 07:43 AM

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i still don't get the answer, can you explain the steps please




#4
Feb2813, 09:37 AM

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calculate when displacement are equald=10(t)1/2(a)(t)^{2} , using (t  0.5) in place of t. (This is because the second stone is in the air for 1/2 second less time than the first stone.) Then "FOIL" the (t  0.5)^{2}, multiply out everything in the expression and then collect terms. See what you have then. 



#5
Feb2813, 10:49 AM

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#6
Feb2813, 11:16 AM

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#7
Feb2813, 11:25 AM

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cas the ffirst stone is .5 seconds in air longer than the second stone 



#8
Feb2813, 05:33 PM

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t for the first stone and (t  0.5) for the second stone, OR (t  0.5) for the first stone and t for the second stone, but not both (t + 0.5) and (t  0.5). 



#9
Feb2813, 08:48 PM

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#10
Feb2813, 09:18 PM

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Therefore, the time for the first stone is always 0.5 s more than the time for the second stone. 



#11
Feb2813, 09:21 PM

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for that cas the time for the first one is 5 less than the time for the second stone. so is this correct "(t + 0.5) for the first stone and t for the second stone,"? 



#12
Feb2813, 10:59 PM

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#13
Feb2813, 11:03 PM

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ohh lol, now i get it, thanks :D



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