2 AP physics B questions

  • #1
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ok so here's one: A helicopter is ascending vertically with a speed of 5.50 m/s. At a height of 155 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [Hint: The package's initial speed equals the helicopter's.]

and here's the other: A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 4.8 s later. If the speed of sound is 340 m/s, how high is the cliff? (no idea how to even begin that one...)

Um, i know that in both of them we're supposed to use the quad. formula and in the first one I tried to but webassign told me I was wrong. I got 11.1s i think. The second I don't even know how to start.
If anyone can do them, can you please explain how you got it? I wanna know how to do them so I don't make mistakes later on in the year.
Thanks.
 

Answers and Replies

  • #2
cepheid
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Um, i know that in both of them we're supposed to use the quad. formula and in the first one I tried to but webassign told me I was wrong. I got 11.1s i think.

Can you post your work? Hint: what maximum height does the package reach before it stops and begins falling? How long does it take something in free fall to fall from a given height?

The second I don't even know how to start.

Here are the two things that have to happen before you hear the sound:

1. The rock actually reaches the bottom.
2. The sound of the impact travels back up to the observer.

You have to figure out how long each of these two things takes, and use the fact that their sum must be 4.8 s in order to find the height of the cliff (which is the distance travelled).
 
  • #3
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ok so for the first one I didn't really know what I was doing so I just followed what the textbook did in an example:

155m= 0 + (5.5m/s)t + 1/2(-9.81m/s^2)t^2
and then the book somehow made that into a quadratic formula and so I did

t= 5.5m/s +- (the square root of...) (5.5m/s)^2 - 4(4.90m/s^2)(-155m)/2(4.90m/s^2)
and I got 2 answers, one was 0.something and the other was 11.1 which didn't work.

and for the second one I get what you're saying. I thought the textbook was telling me like someone out in the woods heard it 4.8s later which confused me. lemme see if I can go try that one now. Thanks
 
  • #4
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For the first one, you are starting at a height of 155m initially.

I've done that problem with the sound thing. You have to assume the sound travels back in a line. I don't know why, when I did the problem like a year ago I was completely dumb-founded by this assumption
 
  • #5
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so basically for the second one, what I have so far is t + t'=4.8s no idea how to get either. I tried isolating them but you still have 2 unknown variables and I tried using some formulas to get t but still you have too many unknowns
 
  • #6
PeterO
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ok so for the first one I didn't really know what I was doing so I just followed what the textbook did in an example:

155m= 0 + (5.5m/s)t + 1/2(-9.81m/s^2)t^2

In this expression, have used 5.5 m/s asa the initial vel - thus saying up is positive

You used -9.81 for g good, the acceleration is down so you said negative

BUT you said the final displacement was 155 - but it ends up below where is started, so that should be -155
 
  • #7
PeterO
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so basically for the second one, what I have so far is t + t'=4.8s no idea how to get either. I tried isolating them but you still have 2 unknown variables and I tried using some formulas to get t but still you have too many unknowns

because the stone was dropped, the time taken to fall is related to the height by

h = 1/2 * 9.81 * t^2 ##

The sound comes back up at constant speed of 340 m/s, so

t' = h / 340

and as you said t = t' = 4.8

That's three equations in three unknowns - solve simultaneously.

[##the usual s = ut + 1/2 8at^2 becomes simple when u = 0]
 
  • #8
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and as you said t = t' = 4.8

wait so the time I thought the total time took 4.8s. like it took 4.8s from when he threw it for the sound to comeback. what I think you're saying is it took 4.8s to hit the water and 4.8s for the sound to come back....or did you accidentally not hit your shift key?
 
  • #9
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also is the 9.81 supposed to be negative since it's gravity?
 
  • #10
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h = 1/2 * 9.81 * t^2 ##

did that and got -113.1264 + 47.136*t - 4.91*t^2

rearranged it in order of degree and then set up my quad. equation and got 0 in the numerator...this problem has pretty much been my morning. i need serious help right now
 
  • #11
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UPDATE: i got the first one, just made an order of operations mistake. the second one is still getting me
 
  • #12
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so I went to my friend's house to work with him on the other one and after 2-3 hours of work, we finally got it. Thanks everyone. kinda crazy how im having this much troubla and school hasnt even started yet....
 
  • #13
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dy=vt + 1/2gt^2
155meters=-5.5t - 4.905t^2
4.905t^2 + 5.5t + 155=0 and solve for t using quadratic forumla.

You should probably draw out the second part. Think about it though...if the speed of sound is
340m/s and the rock was dropped (not thrown), then wouldn't the speed of sound be relavent to straight back up the cliff? well if you think the answer is yes, then you are correct. So, the answer simply lies linearly since you already know how much time it took to go back.... Vyt=height or dy and gravity is not accounted for in this case, because the speed of sound is constant and isn't well shouldnt be affected by gravity. Unless I'm wrong, but I should know enough to be correct.
 
  • #14
318
0
dy=vt + 1/2gt^2
155meters=-5.5t - 4.905t^2
4.905t^2 + 5.5t + 155=0 and solve for t using quadratic forumla.

You should probably draw out the second part. Think about it though...if the speed of sound is
340m/s and the rock was dropped (not thrown), then wouldn't the speed of sound be relavent to straight back up the cliff? well if you think the answer is yes, then you are correct. So, the answer simply lies linearly since you already know how much time it took to go back.... Vyt=height or dy and gravity is not accounted for in this case, because the speed of sound is constant and isn't well shouldnt be affected by gravity. Unless I'm wrong, but I should know enough to be correct.

If this is affected by gravity, which kind of makes sense since the air molecules that the sound waves bounce off of have mass and are affected by gravity, you'd do the following
dy=340t + 4.905t^2, Vy=0=340 - 9.81t (to find how much time it takes to reach top)
t=340/9.81 plug this back into dy equation to get height wait... the problem says that it took 4.8 seconds to reach to the top. so...you dont need to worry about finding time. just plug in 4.8s in the dy equation. actually the equation about finding time seems to prove that based on the given time, that sound wave isn't affected by gravity, since 340 does not equal to 9.81(4.8 seconds).

And if you wanna have fun with the problem, you can even find out the properties of the wave where f=frequency of wave, lamda=wave length, T=time it took to get to the top if you want
340=(f)(lambda)
340/(lambda)=(f)
lamda/340=T
 
  • #15
PeterO
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wait so the time I thought the total time took 4.8s. like it took 4.8s from when he threw it for the sound to comeback. what I think you're saying is it took 4.8s to hit the water and 4.8s for the sound to come back....or did you accidentally not hit your shift key?

Yes missed the shift key t + t' = 4.8. Sorry
 
  • #16
PeterO
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also is the 9.81 supposed to be negative since it's gravity?

Since the displacement and acceleration are both down in this case, you can call down positive.
 

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