Physics Problem

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Ok, I've been working on this one nearly all day.

A bored physics student drops a watermelon from the top of a building, 2.50s later he hears the sound of the watermelon hitting the ground. The speed of sound (as given in this problem) is 340m/s. What is the hieght of the building? (ignore air resistance)


Anyone care to help point me in the correct direction?

so far i have:
2.5s= (time for fall)+(time for soundtravel)

and some manipulated kinematics equations

Acceleration due to gravity = 9.8m per second squared.

yeah... what do I do with that? I kind of already knew that in regards to the accel in the kinematics equations dealing with the watermelon's fall... now all I need is a set up to find the time for either the fall or the time it takes for the sound to reach his ears once the watermelon has generated the sound. or the final velocity when the watermelon hits the ground. or even better, the hieght of the building...

what you know?

time for sound to reach student (from the landing spot) = 2.5 seconds

speed of sound v = 340 m/s (distance over time)

should be able to figure out how many metres traveled in 2.5 seconds using this, right?

ie h (height) in metres = 2.5s * 340 m/s = 850metres (a tall building in my book)

trying to figure out what happened on the way down is a bit of a red herring.

Note: I could have forgotten something important.

edit: aw crap - I have forgotton something important.

Ok I got some sort of decent formula for height of building based on gravity and time

http://www.krysstal.com/formulas.html

g =9.8m/s/s
height = 1/2 * (g * tfall * tfall)

where tfall = time for fall.

tfall + tsound = 2.5
so tsound = 2.5 - tfall

using sound:
height = speed sound * time sound travelled the distance of height.
height = 340m/s * (2.5 - tfall)

so 2h = ( 1/2 * (g * tfall * tfall) ) + (340m/s * (2.5 - tfall) )

so using that incredibly grotty thing and my bodgy arithmetic
I got this even grottier thing

h = 2.45 * (tfall*tfall) - 170*tfall + 425

now there should be some backwards way of substituting back in the time, after all we know that t fall is between 0 and 2.5 and t sound is also between 0 and 2.5

so maybe if we work with the two formulas for height like

( 1/2 * (g * tfall * tfall) ) = 340m/s * (2.5 - tfall)

we could get a number for tfall.

I can't believe how crap I am at this now.

Depending what you use for g

I've ended up with

t fall = 2.417
t sound =0.083
height = 28.22m

All a bit fuzzy really. Our physics teachers never made the numbers as mad as this.

well, the h you found is within the range for the correct answer. according to the book its 28.6m but you are pretty much dead on..

thanks so much for the help, this thing has been bugging me all day.

Unless I have made a mistake (as is so very often the case), the watermelon took around 2.416 seconds to reach the ground and immediately became vegetarian road pizza. The sound took around 84.17 milliseconds to inform the aformentioned college student of this sad fact. The building is around 28.62 meters high, or 29.60 meters if you round gravity to two significant figures.

- M

Maka Ra

I knew there was some reason I run away from physics problems like these. That looks very elegant - I have nothing on my computer that makes beautiful symbols like that.

And once I'd worked out the quadratic equation with time for fall - I couldn't figure out how to get the number out. I knew it was possible but it wouldn't unravel. Rusty brain connections I suppose.

I did my usual thing of making lots of huge aproximations because I am lazy. If you do it the lazy way you find out that the time it takes for the sound to travel back is pretty small, so I then felt justified in ignoring it. I got a final answer of 30m. not very precise, but close enough given that I used g = 10. Now the student knows that jumping off after his beloved melon would be a bad idea.

Just out of curiousity what level of physics is everyone at?

Hi everybody,

t = time in seconds,
h = height of building in meters,
v(t) = velocity of watermellon at time t,
z(t) = position of watermellon at time t.
f = total time to fall in seconds,
s = time for sound to return in seconds,
f+s = 2.495 to 2.505 seconds
340 = speed of sound in meters/second.
g = 9.80665 meters/second^2
ignore air resistance.

v(t) = gt
z(t) = 1/2 g t^2
f = t, such that z(t) = h,
or easier,
h = z(f) = 1/2 g f^2.
Eliminate h:
s = h / (340 m/s) = 1/2 g f^2 / 340 = (g/680) f^2
g/680 f^2 + f + c = 0
Quadratic equation:
a f^2 + b f + c = 0
a = g/680
b = 1
c = -2.495 or -2.5 or -2.505
f = (-b + sqrt(b^2 -4ac))/2a
f = 2.4112 or 2.4158 or 2.4205 seconds to fall
h = 28.51 or 28.62 or 28.73 meters high, i.e.,
h = 28.62 +/- 0.11 meters,
given time to hear sound was 2.50 +/- 0.005 seconds.

-- Archimerged (Arc)

archimerged

dam n, don't even need to measure it to get three different answers.

It's not that complex. You guys are over-thinking it. In the 1st second, it falls 9.8 metres. In the 2nd second, it falls (9.+(9. metres, having gained an extra 9.8 metres over that second. Then from second 2.0 to 2.5, it falls 19.6+(9.8/2) metres.

Second 1: 9.8 metres.
Second 2: 19.6 metres.
Second 2.5: 24.5 metres.

Answer: The building is 24.5 metres high.

What some people forget is although gravity causes an object to travel at a speed of 9.8 m/s after one second, it has not travelled 9.8m because after 1/2 second it would have only been travelling at about 3m/s.
I have found a rough guide to work things out is to say that the object has travelled 5m after 1s, (then add 10 to the speed every second) 20m after 2s, 45m after 3s, 80m after 4s, ect.
At 2 1/2 seconds it would have have travelled a distance of about 30m.