Projectile motion in one dimension

[Random Variable]

This supposedly simple problem is giving me nothing but fits.

Milton drops a 56K dial-up modem over the side of a cliff. Exactly four seconds after dropping the modem he hears the sound of it hitting the ground below. Assuming that the speed of sound is 340m/s, what is the height of the cliff? Ignore air resistance/friction.

 

[Yossarian]

n/m, give me a minute

 

[Ameesh]

is milton on earth?



wooo WOOO!

 

[Random Variable]

Originally posted by: PipBoy
you're kidding right? is there some trick here?

Then what's the answer?

 

[Random Variable]

Originally posted by: Ameesh
is milton on earth?



wooo WOOO!

Yep.

 

[Heisenberg]

The total time is takes for him to hear the sound will be the time of fall plus the time it takes for the sound to reach him. You'll then also have an expression for the distance in terms of the fall time and the time for the sound to reach him. Two equations and two unknowns, which should be solveable. I think that will work - gimme a minute to really think about it.

 

[NakaNaka]

Am I the only one who thinks you need a little trial and error? Because you have d=1/2at^2 (because theres no inital velocity. You have two variables, the time and the distance. You can't calculate how long it takes the sound to travel back up unless you know the time or the displacement. Or so it seems to me.

 

[notfred]

Distance traveled of modem (DM) = 4.9t^2, where t is time in seconds.

Distance travelled of sound (DS) = 340t

At t = 4, DM = DS

Is that how you set the problem up?

 

[FenrisUlf]

For dropped object:
distance = 1/2 * acceleration * time^2

For sound
distance = time * velocity of sound

distances are equal, acceration = 9.8 m/s^2

time of dropped object (t1) + time of sound (t2) = 4
t1 = 4 - t2

4.9 * (4 - t2)^2 = t2 * 340

solve for t2 then put in second equation to get distance.

 

[Random Variable]

Originally posted by: notfred
Distance traveled of modem (DM) = 4.9t^2, where t is time in seconds.

Distance travelled of sound (DS) = 340t

At t = 4, DM = DS

Is that how you set the problem up?

DM does not equal DS when t=4. In purely mathematical terms, when t=4, the modem is below the ground.

 

[NakaNaka]

But there are different times. There is the time it takes for the modem to fall and the time it takes for the sound to come back up. T1+T2=4.

I dunno - thats just how I see it.

 

[Heisenberg]

Originally posted by: FenrisUlf
For dropped object:
distance = 1/2 * acceleration * time^2

For sound
distance = time * velocity of sound

distances are equal, acceration = 9.8 m/s^2

time of dropped object (t1) + time of sound (t2) = 4
t1 = 4 - t2

4.9 * (4 - t2)^2 = t2 * 340

solve for t2 then put in second equation to get distance.

^^^ He got it. That's more or less what I was saying above.

 

[Random Variable]

Originally posted by: NakaNaka
But there are different times. There is the time it takes for the modem to fall and the time it takes for the sound to come back up. T1+T2=4.

I dunno - thats just how I see it.

That's why I'm so confused.

 

[MichaelD]

42. The answer is ALWAYS 42. EVen threads about colors or food or religion. 42.

 

[notfred]

Originally posted by: Vespasian

Originally posted by: notfred
Distance traveled of modem (DM) = 4.9t^2, where t is time in seconds.

Distance travelled of sound (DS) = 340t

At t = 4, DM = DS

Is that how you set the problem up?

DM does not equal DS when t=4.

Yes they do. Let's jsut assume the modem fell 100meters. The modem went down 100 meters, the sound went up 100 meters.

 

[FenrisUlf]

You have two times because there are two distinct events that happen one after the other. The first is the modem falling to the bottom. T1. The second is the sound coming from that event back to Milton, T2. You know that the combined time is 4 seconds, therefore T1 + T2 = 4

 

[Random Variable]

Originally posted by: notfred

Originally posted by: Vespasian

Originally posted by: notfred
Distance traveled of modem (DM) = 4.9t^2, where t is time in seconds.

Distance travelled of sound (DS) = 340t

At t = 4, DM = DS

Is that how you set the problem up?

DM does not equal DS when t=4.

Yes they do. Let's jsut assume the modem fell 100meters. The modem went down 100 meters, the sound went up 100 meters.

But the modem hits the ground before t=4.

 

[blahblah99]

distance = 1/2*acceleration*time*time (assuming the object is initially at rest which in this case it is.). time = time it took for modem to travel distance.

You know acceleration. You need to find distance and time it took the modem to fall.

You also know distance = velocity * T2, where T2 is (4seconds - time).

velocity = 340 m/s
acceleration = 9.8 m/s/s
equate the two equation and solve for time, then plug it back into one of the equations and get distance.

 

[notfred]

Originally posted by: Vespasian

Originally posted by: notfred

Originally posted by: Vespasian

Originally posted by: notfred
Distance traveled of modem (DM) = 4.9t^2, where t is time in seconds.

Distance travelled of sound (DS) = 340t

At t = 4, DM = DS

Is that how you set the problem up?

DM does not equal DS when t=4.

Yes they do. Let's jsut assume the modem fell 100meters. The modem went down 100 meters, the sound went up 100 meters.

But the modem hits the ground before t=4.

I Guess you think D in my proplem stands for time rather than distance?

 

[rgwalt]

Originally posted by: FenrisUlf
For dropped object:
distance = 1/2 * acceleration * time^2

For sound
distance = time * velocity of sound

distances are equal, acceration = 9.8 m/s^2

time of dropped object (t1) + time of sound (t2) = 4
t1 = 4 - t2

4.9 * (4 - t2)^2 = t2 * 340

solve for t2 then put in second equation to get distance.

oWnEd

Ryan

 

[Random Variable]

Originally posted by: FenrisUlf
For dropped object:
distance = 1/2 * acceleration * time^2

For sound
distance = time * velocity of sound

distances are equal, acceration = 9.8 m/s^2

time of dropped object (t1) + time of sound (t2) = 4
t1 = 4 - t2

4.9 * (4 - t2)^2 = t2 * 340

solve for t2 then put in second equation to get distance.

Yeah, you're right. I'm an idiot.

 

[FenrisUlf]

Originally posted by: Vespasian

Originally posted by: notfred

Originally posted by: Vespasian

Originally posted by: notfred
Distance traveled of modem (DM) = 4.9t^2, where t is time in seconds.

Distance travelled of sound (DS) = 340t

At t = 4, DM = DS

Is that how you set the problem up?

DM does not equal DS when t=4.

Yes they do. Let's jsut assume the modem fell 100meters. The modem went down 100 meters, the sound went up 100 meters.

But the modem hits the ground before t=4.

There are two equations and two times for two events - modem going down and sound going up. The total time is four seconds.

 

[notfred]

I get the cliff being 70.38meteres high.

 

[Yossarian]

Originally posted by: notfred
I get the cliff being 70.38meteres high.

yep, close enough

 

[FenrisUlf]

Originally posted by: notfred
I get the cliff being 70.38meteres high.

Ding Ding! You get a gold star!