Question
A particle is projected vertically upward in a constant gravitational field with
an initial speed of v0 . Show that if there is a retarding force proportional to
the square of the speed, the speed of the particle when it returns to its initial
position is
v0 vT
q
,
v02 + vT2
where vT is the terminal speed.
Answer
Do this question in 2 parts.
Upward motion
y
6
•
?
mg
?
mk ẏ 2
Using Newton’s 2nd law gives:
mÿ = −mk ẏ 2 − mg
dv
dv
dv
= −kv 2 − g Put
=v
dt
dt
dy
Z
v dv
= −y + C
kv 2 + Y
Ã
!
1
kv02 + g
whence y =
ln
,
2k
kv 2 + g
where v0 is the initial upwards velocity.
At the highest point v =Ã0.
!
k02 + g
1
Height attained =
ln
2k
g
Downward motion
mk ẏ 2
6
•
?
mg
Using Newton’s 2nd law gives:
1
ÿ = k ẏ 2 − g
Ã
Ã
!
!
g − kv 2
kv02 + g
1
1
ln
ln
y =
+
, (∗)
2k
g
2k
g
where v = 0 at the highest
r point.
g 2
v
Solving for v gives v = v2k+0g
0 k
Terminal velocity occurs when there is nor net force on the mass while the
g
.
particle is falling i.e. mkvt2 = mg ⇒ vt =
k
Therefore v = q
2
v0 vt
v02 + vt2