Kinematics : A Rock Being Dropped off a Cliff

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[itex][/itex]

Homework Statement


A Rock is dropped from a cliff and the sound of it striking the ocean is heard 3 seconds later. If the speed of sound in air is 340m/s, how high is the cliff?


Homework Equations


The linear kinematics of motion equations:

1. [itex]x-x_{0}=v_{0}t+\frac{1}{2}at^2[/itex]
2. [itex]v=v_{0}+at[/itex]
3. [itex]v^2 =v_{0}^2 + 2a(x-x_{0})[/itex]


The Attempt at a Solution


The problem I have is coming up with a mathematical model to describe what is going on here. I cannot think of a way to set-up these equations to give me an answer. There are two main problems about this that I cannot get passed:

1. I don't know the height of the cliff, or how to work it out (the sound comes three seconds after the stone is dropped, not when the stone strikes the ocean, so I cannot use speed = distance/time using 3s and 340m/s to work out the height of the cliff)

2. The time it takes for the rock to hit the ocean is not the same as the time it takes for the sound to reach the top of the cliff from the ocean (so I cannot use the time parameter to solve two equations simultaneously, as well as I cannot seem to find/come up with any model with gives me something like t1 + t2 = 3s)

Could someone please give me a hint on how to evaluate this question? Because I'm majorly stuck :biggrin:
 

Answers and Replies

  • #2
rude man
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1. Assuming height h, how much time t1 does it take for the rock to fall to the ocean?
2. How much time t2 does it take for the sound to reach the thrower after the rock hit the ocean?
3. Do you know t1 + t2?

3 equations, 3 unknowns?
 
  • #3
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You can use T1 + T2 = 3 to substitute into the kinematic equations. For example:T1=3-T2. Figure out an equation that relates the height of the cliff to each individual time and then you can substitute the T1 value above.

I'll get you started:

Use your first equation.
y=Vi*t+1/2at^2 where y is the height of the cliff. Vi=0 because it is dropped.
y=-4.9(T1)^2
 
  • #4
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t=t1+t2 and t=3s and v=340[itex]\frac{m}{s}[/itex]
t1=[itex]\sqrt{\frac{2h}{g}}[/itex] and t2=[itex]\frac{h}{v}[/itex]

therefore t=[itex]\sqrt{\frac{2h}{g}}[/itex] + [itex]\frac{h}{v}[/itex]
t2=[itex]\frac{2h}{g}[/itex]+ [itex]\frac{h^2}{v^2}[/itex]
t2gv2=2hv2+h2g you know t,g and v
and then the quadratic formula
 
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  • #5
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y=-4.9(T1)^2
If I re-arrange this to make T1 the subject, wouldn't I then get a negative root?

EDIT: here is a full attempt, I know it is wrong. But I cannot think of where I made a mistake.
 

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  • #6
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I think you made mistake at 36d2/3402
 
Last edited:
  • #7
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Ahh, ok...
 

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