Solving Kinematics Problem: Two Stones Dropped from 60m Cliff

AI Thread Summary
Two stones are dropped from a 60-meter cliff, with the second stone released 1.6 seconds after the first. The problem involves calculating the distance of the second stone from the top of the cliff when the separation between the two stones reaches 36 meters. The correct answer to the problem is 10.9 meters. Participants in the discussion emphasize the importance of posting in the appropriate forum section and encourage preparation for the upcoming quiz. Overall, the conversation centers on solving a kinematics problem related to free-fall motion.
superstorebug
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I'm fairly new with this so I hope I've posted at the right place!


I don't understand how to go about doing the following kinematics problem:

Two stones are dropped from the edge of a 60-m cliff, the second stone 1.6s after the first. How far below the top of the cliff is the second stone when the separation between the two stones is 36 m?

I hope you can help because I have a quiz on this stuff tomorrow! Thanks in advance.

ANy help will be greatly appreciated.
 
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Oh the answer is 10.9 meters
 
Welcome to physics forums

Good job finding the answer. Next time please check andpost in the right section though >.>

get well rested for that quiz
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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