Physics 3221 Mechanics I
Fall Term 2010
Quiz 2
This is a 20 min. quiz (closed book). There is only one problem.
Problem 1. [5 pts] You are standing next to a building and want to toss your cell phone to
your friend who is leaning out of a third-floor window which is a distance h above the ground.
Assume that the air drag is proportional to the square of the velocity, i.e. Fr ≡ |F~r | = mkv 2 .
(a) Find the minimum velocity vmin with which you must throw the cell phone vertically
up so that it can reach your friend. Express your answer in terms of h, k and g.
(b) Check the validity of your answer from part (a) by discussing the following limits: k → 0,
g → 0, h → 0, k → ∞, g → ∞, h → ∞.
Solution. (a) Choosing the z axis pointing up, Newton’s law reads
mz̈ = −mg − mk ż 2
(the particle is moving up hence the friction force acts downwards). The problem only involves
the velocity v and position z, thus time is not relevant to this problem and it is convenient to
rewrite Newton’s law in terms of ż(z) = v(z) instead of ż(t):
z̈ = v̇ =
dv dz
dv
dv
dv
=
= ż = v
= −(g + kv 2 )
dt
dz dt
dz
dz
and integrate:
Z
vdv
=−
g + kv 2
Z
dz + C
which gives
1
ln(g + kv 2 ) = −z + C.
2k
If the ground level is at z = 0, C is determined by the initial condition v(z = 0) = vmin :
C=
1
2
ln(g + kvmin
)
2k
which gives
g + kv 2
ln
2
g + kvmin
!
= −2kz.
vmin corresponds to the case when v = 0 at z = h:
g
ln
2
g + kvmin
Solving for vmin ,
vmin =
r
!
= −2kh.
g +2kh
(e
− 1)
k
1
Solution. (b)
1) No friction (k = 0). Expand for small k:
vmin =
r
q
g
(1 + 2kh + o(k 2 ) − 1) = 2gh,
k
which is the usual formula in the absence of friction.
2) No gravity (g = 0). Then vmin → 0, i.e. you throw infinitely slow.
3) No elevation (h = 0). Then vmin → 0, i.e. no need to throw.
4) Infinite friction (k = ∞). Then vmin ∼
e∞
∞
→ ∞ and you have to throw infinitely hard.
5) Infinite gravity (g = ∞). Then vmin → ∞ and you have to throw infinitely hard.
6) Infinitely tall building (h = ∞). Then vmin → ∞ and you have to throw infinitely hard.
2