You’re hanging out by a campfire. You notice a tiny piece of fly ash that is eje

Question

You’re hanging out by a campfire. You notice a tiny piece of fly ash that is ejected from the fire vertically upward with initial speed v0. Since physicists tend to love to think about casual observations like this more formally, you decide to try to predict if the ash will reach tree-level. Let’s measure the position of the fly ash, y from the point of release, taking vertically upward as the positive direction.

1. Clearly the physics of “flying ash” could be very complicated! But, let’s start with the simplest possible model and go from there. If you consider ash to be a point particle of mass m acting only under gravity, what is the equation of motion? What is the maximum height (in terms of v0 and g)?

2. The next simplest model might be to also include simple air drag. Knowing that ash particles are extremely tiny, and not particularly fast moving to start with, let’s include a linear air drag in the equation of motion. Write equation of motion (i.e., Newton’s second law), which includes a linear air drag.

3. Now, using this model, find the time for the fly ash to reach its highest point and its position ymax at that time. (Note that velocity vy(t) and position y(t) are derived in the text. However, you should be careful, because in the text, Taylor defines +y to be the other way, so watch your minus signs! In this question, please use upward direction to be positive.)

I understand that it looks like more than just 1 question, but its just breaking the 1 question into easier parts to walk through it. Please help! Thank you!

Explanation / Answer

given flyash piece has initial upward velocity vo

1. considering it to be a point mass m acting under ghravity only

then height y above the surface at timet is given by

y = vot - 0.5gt^2

2. for linear drag

from newtons second law

my" = mg - kv

v = y'

hence

my" = mg - ky'

3. mdv/dt = mg - kv

dv/(mg/k - v) = kdt/m

integrating

ln((mg/k - v)/(mg/k - vo)) = -kt/m

v = mg/k - (mg/k - vo)exp(-kt/m) = dy/dt

integrating

mgt/k + m(mg/k - vo)[exp(-kt/m) - 1]/k = y

at maximum height

v = 0

t = mln(1 - vo*k/mg)/k

hence ymax = m^2g*ln(1 - vo*k/mg)/k^2 + m[vo]/k

ymax = mvo/k - m(vo/k + vo^2/mg) = vo^2/g

4. as k -> 0

ymax = m[vo]/k

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