Trouble with Two Physics Problems: Speed of Sound and Acceleration

In summary: So it would be \sqrt{\frac{h}{4.9}} + \frac{h}{340} = 3.4. Then you can solve for h. As for #2, you can use the equation d=vt to find the distance the motorist traveled, and then use the equation d=0.5at^2 to find the distance the cop traveled. Set these two equal and solve for t, which will give you the time it takes to catch the speeder. Then you can use the equation v=u+at to find the velocity of the cop (where u is the initial velocity of the cop, which is 0). In summary, for the first question, the height of the cliff can be
  • #1
cscott
782
1
I'm reviewing for a test on everything I learned last year and I'm having trouble with these two questions:

1. A rock is dropped from a sea cliff and the sound is heard 3.4s later. Speed of sound is 340 m/s, how high is the cliff?

2. A motorist passes a cop at 120 km/h. The cop begins immediately accelerating at 10 km/h/s. How much time will it take to catch the speeder and how fast will the officer be travelling?

I can't get my equations right so I have to many unknowns... the sad thing is I've seen both of these questions before... :\
 
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  • #2
cscott said:
...

I can't get my equations right so I have to many unknowns... the sad thing is I've seen both of these questions before... :\

Isolate the known constants and unknown variables. What types of constants are present in the first question (time for something to go a certain distance, distance after uniform velocity, distance after uniform acceleration, etc.)? What is the unknown ? What kinematics equation(s) have only these types of variables present ?
 
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  • #3
1) As the rock is falling downward, it's height is given by h(t)= -4.9t2+ h0 where x0 is the height of the cliff. How long will it take to hit the bottom, as a function of h? Of course, the sound takes time vs/h (vs is the speed of sound (340 m/s) and h is the (unknown) height of the cliff) to come back to your ear. The sum of those two formulas is, according to this problem, 3.4 sec. Solve for h.

1) The distance moved by the motorist, in miles where t is in hours, is sm(t)= 120 t. The distance moved by the policeman is sp(t)= (1/2)(10 km/h/s) t2. Set those equal and solve for t. You might want to change "10 km/h/s to km/h2 to keep everything in the same units.
 
  • #4
HallsofIvy said:
1) As the rock is falling downward, it's height is given by h(t)= -4.9t2+ h0 where x0 is the height of the cliff. How long will it take to hit the bottom, as a function of h? Of course, the sound takes time vs/h (vs is the speed of sound (340 m/s) and h is the (unknown) height of the cliff) to come back to your ear. The sum of those two formulas is, according to this problem, 3.4 sec. Solve for h.

1) The distance moved by the motorist, in miles where t is in hours, is sm(t)= 120 t. The distance moved by the policeman is sp(t)= (1/2)(10 km/h/s) t2. Set those equal and solve for t. You might want to change "10 km/h/s to km/h2 to keep everything in the same units.


I'm probably just tired and this isn't making sense to me, but wouldn't the time the sound takes to reach your ear be h/vs?

h(t) = 0 = -4.9t2 + h0

...solve for t and then add (as HallsofIvy said).

Edit: I don't know why my superscript isn't working, but you get the point :smile:
 
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  • #5
Maybe I'm tired as well, but for #1 do I end up with [tex]\sqrt{\frac{h}{-4.9} }+ \frac{h}{340} = 3.4[/tex]

When I try to do this I get some crazy quadratic... at least I see I was on track.

Thanks for your help.
 
  • #6
cscott said:
Maybe I'm tired as well, but for #1 do I end up with [tex]\sqrt{\frac{h}{-4.9} }+ \frac{h}{340} = 3.4[/tex]

When I try to do this I get some crazy quadratic... at least I see I was on track.

Thanks for your help.

That should be 4.9 (not -4.9).
 
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