Solving for distance fallen when object is already moving when time begins

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califauna

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Hi,

The question posed is:

A person looking out of a window of a tall building sees a bucket fly past the window at 30 m/s, then hears the bucket hit the ground 8 seconds later. At what altitude is the observer located? Assume negligible air resistance and speed of sound at 340 m/s.

The problem I am having is that the sound takes an unknown time to arrive at the observer after hitting the ground.


Using the common kinematic equations and substitution, I have built a couple of formulas:

attachment.php



What are the steps using my equation, if correct? I haven't managed to do it.

Or, how do I solve it after setting equation#1 equal to velocity * (total time - time for sound to reach observer)?

I have found various equations for solving similar penny/stone in the well problems , but they all assume a starting velocity of zero.

As an additional question, what area of algebra should I practice here, in order to solve equations like this?

Thanks!
 

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Perhaps ..

califauna said:
Hi,

The question posed is:

A person looking out of a window of a tall building sees a bucket fly past the window at 30 m/s, then hears the bucket hit the ground 8 seconds later. At what altitude is the observer located? Assume negligible air resistance and speed of sound at 340 m/s.

The problem I am having is that the sound takes an unknown time to arrive at the observer after hitting the ground.


Using the common kinematic equations and substitution, I have built a couple of formulas:

attachment.php



What are the steps using my equation, if correct? I haven't managed to do it.

Or, how do I solve it after setting equation#1 equal to velocity * (total time - time for sound to reach observer)?

I have found various equations for solving similar penny/stone in the well problems , but they all assume a starting velocity of zero.

As an additional question, what area of algebra should I practice here, in order to solve equations like this?

Thanks!
Click to expand...

Your factor (8 seconds - (t seconds)/340 meters) is subtracting seconds/meter from seconds, i.e. apples from oranges.

The time delay in the sound travelling from ground to ear is (d meters/340 m/s), overall, leaving you with a single equation that is a function of d to solve.
 
califauna said:
Hi,

The question posed is:

A person looking out of a window of a tall building sees a bucket fly past the window at 30 m/s, then hears the bucket hit the ground 8 seconds later. At what altitude is the observer located? Assume negligible air resistance and speed of sound at 340 m/s.

The problem I am having is that the sound takes an unknown time to arrive at the observer after hitting the ground.


Using the common kinematic equations and substitution, I have built a couple of formulas:

attachment.php



What are the steps using my equation, if correct? I haven't managed to do it.

Or, how do I solve it after setting equation#1 equal to velocity * (total time - time for sound to reach observer)?

I have found various equations for solving similar penny/stone in the well problems , but they all assume a starting velocity of zero.

As an additional question, what area of algebra should I practice here, in order to solve equations like this?

Thanks!
Click to expand...
The way I would start to approach the problem is somewhat different than yours:
h = height above ground
d(t) = distance bucket falls as a function of time t measure from the time the bucket passes the window:
d(t) = h - 30 t - 5 t2
from simple ballistic equation. Solve for d(t0) = 0 as a function of h, i.e. at t0(h), d(t0)=0, which is the solution of a quadratic equation. After bucket hits the ground, the time it takes to reach height h is
t1(h) = h/340
Let f(h) be the function
f(h) = t0(h) + t1(h) - 8
Find the zeros of f.

Actually, right off hand, it look like a quadratic so it should not be bad.

EDIT: BTW, I didn't mean f(h) was quadratic. I meant the resulting equation in solving for the zeros of f could be expressed as a simple quadratic equation.
 
Last edited:
Picture it

In problems involving constant acceleration, it is best to draw a velocity graph (verses time)
Why?
1) The velocity is a straight line.
2) The acceleration is the slope of that line.
3) The distance is the Area under that line.

Let
v = initial velocity
vf = final velocity

then
(vf-v)/t = a = slope
vf = at+v
attachment.php
 

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Dale10101 said:
Your factor (8 seconds - (t seconds)/340 meters) is subtracting seconds/meter from seconds, i.e. apples from oranges.

The time delay in the sound travelling from ground to ear is (d meters/340 m/s), overall, leaving you with a single equation that is a function of d to solve.
Click to expand...

Ah yes thanks, I should have put meters in there. Ok Im going to try again to see if I can solve for d.
 
Bob Brown MSEE said:
In problems involving constant acceleration, it is best to draw a velocity graph (verses time)
Why?
1) The velocity is a straight line.
2) The acceleration is the slope of that line.
3) The distance is the Area under that line.

Let
v = initial velocity
vf = final velocity

then
(vf-v)/t = a = slope
vf = at+v
attachment.php
Click to expand...

But the problem is that I dont know know the value of t(total time to reachobservers ears). I can see how your graphs help understand the final velocity but Im trying to solve for the distance, and the second time component mixed into the total time has me thrown. Im not ure how I can calculate based on your methods suggested.
 
Denis said:
Unless I missed something, seems to be this easy:

d = 9.8t : falling window to pavement
d = 340(8-t) : sound travelling back up to window

9.8t = 340(8-t)
t = ~7.776

d = 9.8 * 7.776 = ~76.2
Click to expand...
Denis,

I was under the impression that we had an initial velocity of 30 m/s straight down and and an acceleration downward of 10 m/s2, I rounded the 9.807 to a whole number but you could just change my 5 to 4.9035 or round that off to 4.9. The simple ballistic velocity and distance equations and for constant acceleration straight down [no x component] are
v(t) = v0 + a t = -30 - 10 t
d(t) = h - 30 t - 5 t2
if we start time zero when the bucket passes the window and distance is measured from the ground up. Of course, things change when the bucket hits the ground. So the time to hit the ground after passing the window is the smaller positive solution of d(t)=0 or
\(\displaystyle t_0\, =\, \dfrac{-30\, +\, \sqrt{900\, + 20 h}}{10}\)

Now the time taken for the sound starting at zero and reaching height h is
t1 = h/340

The solution to our problem is thus
t = t0 + t1 = 8
or
\(\displaystyle \dfrac{-30\, +\, \sqrt{900\, + 20 h}}{10}\, +\, \dfrac{h}{340}\, =\, 8\)

You can manipulate that into a quadratic equation for h and solve for h. The answer is about 429m. If I change the 5 to 4.9035, the answer is h is about 426m.

EDIT: Change dumb mistake.
 
Last edited:
Denis said:
Unless I missed something, seems to be this easy:

d = 9.8t : falling window to pavement
d = 340(8-t) : sound travelling back up to window

9.8t = 340(8-t)
t = ~7.776

d = 9.8 * 7.776 = ~76.2
Click to expand...

Thanks. First of all I should say that my equation is wrong to start with, as pointed out by Dale10101. I think I should have substitued the total time with (8-d/340).

Yes, Im trying to solve for d, and the second time component (when the sound travels to the observer) has me thrown.

Something not right there. The answer is 425.6 meters. The time it took to reach observer is 1.25 seconds. As I dont know howthey got to the answer, I tried solving by using substitution, but I cant solve it for d. (below is answer, in Catalan!)

http://postimg.org/image/mv8rt46gn/
 
Here is answer:

http://postimg.org/image/mv8rt46gn/

As I dont know how they got to the answer, I tried solving it using substitution then solving for d by jiggling the equation around to solve for D, but I havent managed to do it so far. Before other methods please let me know if I can solve it this way so I know where Im going wrong.
 
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