Kinematics : A Rock Being Dropped off a Cliff

1. Mar 8, 2013

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
The linear kinematics of motion equations:

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

3. 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

2. Mar 8, 2013

rude man

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. Mar 8, 2013

KHANV1CT

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:

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. Mar 9, 2013

lep11

t=t1+t2 and t=3s and v=340$\frac{m}{s}$
t1=$\sqrt{\frac{2h}{g}}$ and t2=$\frac{h}{v}$

therefore t=$\sqrt{\frac{2h}{g}}$ + $\frac{h}{v}$
t2=$\frac{2h}{g}$+ $\frac{h^2}{v^2}$
t2gv2=2hv2+h2g you know t,g and v

Last edited: Mar 9, 2013
5. Mar 10, 2013

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.

Attached Files:

• The rock falling off a cliff.pdf
File size:
228.3 KB
Views:
129
Last edited: Mar 10, 2013
6. Mar 10, 2013

lep11

I think you made mistake at 36d2/3402

Last edited: Mar 10, 2013
7. Mar 10, 2013