479 Horizontal Motion: The horizontal component of velocity ... - Wuala

91962_05_R1_p0479-0512 6/5/09 3:56 PM Page 509
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R1–46. A particle of mass m is fired at an angle u 0 with a
velocity v 0 in a liquid that develops a drag resistance
F = -kv, where k is a constant. Determine the maximum
or terminal speed reached by the particle.
y
Equation of Motion: Applying Eq. 13–7, we have
: + ©F x = ma x ; -ky x = ma x a x = - k m y x
[1]
O
v 0
0 u
x
+ c ©F y = ma y ; -mg - ky y = ma y a y = -g - k m y y
[2]
However, , , and a y = d2 y
a y y = dy
x = d2 x
y x = dx
. Substitute these values into
dt dt 2 dt
dt 2
Eqs. [1] and [2], we have
d 2 x
[3]
dt 2 + k dx
m dt = 0
d 2 y
[4]
dt 2 + k dy
m dt = -g
The solution for the differential equation, Eq. [3], is in the form of
Thus,
x = C 1 e - k m t + C 2
x # C 1 k
= -
k m
e- m t
[5]
[6]
However, at t = 0, x = 0 and x # = y 0 cos u 0 . Substitute these values into Eqs. [5] and
m
[6], one obtains C and C 2 = m 1 = -
. Substitute C 1
into Eq. [6]
k y 0 cos u
k y 0 cos u 0
0
and rearrange. This yields
x # = e - k m t (y 0 cos u 0 )
The solution for the differential equation, Eq. [4], is in the form of
y = C 3 e - k m t + C 4 - mg
k t
[7]
[8]
Thus,
y # C 3 k
= -
k m
e- m t - mg
k
[9]
However, at t = 0, y = 0 and y # = y 0 sin u 0 . Substitute these values into Eq. [8] and
[9], one obtains C m
and C 4 = m .
k ay 0 sin u 0 + mg
3 = -
k ay 0 sin u 0 + mg
k b
k b
Substitute C 3
into Eq. [9] and rearrange. This yields
y # = e - k m t ay 0 sin u 0 + mg
k b - mg
k
[10]
For the particle to achieve terminal speed, t : q and e - k m t : 0. When this happen,
from Eqs. [7] and [10], and y y = y # mg
y x = x # = 0
= - . Thus,
k
y max = 2y 2 x + y 2 y = 0 2 mg
2
+ a - C k b = mg
k
Ans.
509