Height of a Cliff (Only time Given)

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To determine the height of the cliff, the time delay of 3.4 seconds includes both the rock's fall and the sound's travel time. The rock's fall time is less than 3.4 seconds, while the sound takes time to travel back up at 340 m/s. A system of two equations should be set up: one for the distance the rock falls and another for the distance the sound travels. The equations must relate the heights and the times involved. This approach will yield a more accurate calculation of the cliff's height.
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Homework Statement


A person throws a rock off a cliff and hears the splash below 3.4 seconds later. If we know the speed of sound is equal to 340m/s how high is the cliff to the nearest metre?


Homework Equations


I've been trying to use, r = ut + 1/2at2 and
v2 = u2 +2ar and
v= u + at


The Attempt at a Solution


I know that the height of the cliff should be around 56 m as:

r = 0 + 4.9(3.4)2

But i don't know how to factor in the time delay of the sound :(

The answer is: (Spoiler)
52 m

Thanks
 
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PhysGoat said:
I know that the height of the cliff should be around 56 m as:

r = 0 + 4.9(3.4)2

But i don't know how to factor in the time delay of the sound :(

Welcome to PF!

Hint: it does not take 3.4 seconds for the rock to fall, it takes slightly less than that.
 
Nice problem, i like it. You should set up a system of 2 equations. One will be the distance when the rock falls down and the other the distance when the sound comes up.
What is the relation between the two equations and what is the relation between the two times?
 
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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