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

91962_05_R1_p0479-0512 6/5/09 3:56 PM Page 510
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
R1–47. A projectile of mass m is fired into a liquid at an
y
angle u 0 with an initial velocity v 0 as shown. If the liquid
develops a friction or drag resistance on the projectile
which is proportional to its velocity, i.e., F = -kv, where k
is a constant, determine the x and y components of its
position at any instant. Also, what is the maximum distance
x max that it travels?
Equation of Motion: Applying Eq. 13–7, we have
O
v 0
0 u
x
: + ©F x = ma x ; -ky x = ma x a x = - k m y x
[1]
+ 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
. Substituting 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 . Substituting these values into Eq. [5]
m
and [6], one can obtain C and C 2 = m 1 = -
. Substituting C
k y 0 cos u
k y 0 cos u 0
0
1
and C 2
into Eq. [5] and rearrange yields
x = m k y 0 cos u 0 A1 - e - k m t B
Ans.
When t : q, e - k m t : 0 and x = x max . Then,
x max = m k 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
Ans.
[7]
Thus,
y # C 3 k
= -
k m
e- m t - mg
k
[8]
However, at t = 0, y = 0 and y # = y 0 sin u 0 . Substitute these values into Eq. [7] and
[8], one can obtain 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
and C 4
into Eq. [7] and rearrange yields
y = m k ¢y 0 sin u 0 + mg
k ≤ a1 - e- k m t b - mg
k t
Ans.
510