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PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, 
you are using it without permission. 
171 
 
 
PROBLEM 11.115 
An oscillating garden sprinkler which 
discharges water with an initial velocity 
v0 of 8 m/s is used to water a vegetable 
garden. Determine the distance d to the 
farthest Point B that will be watered and 
the corresponding angle α when (a) the 
vegetables are just beginning to grow, 
(b) the height h of the corn is 1.8 m. 
 
SOLUTION 
First note 0 0
0 0
( ) cos (8 m/s) cos
( ) sin (8 m/s) sin
x
y
v v
v v
α α
α α
= =
= =
 
Horizontal motion. (Uniform) 
 00 ( ) (8 cos )xx v t tα= + = 
At Point B: : (8 cos )x d d tα= = 
or 
8 cosB
d
t
α
= 
Vertical motion. (Uniformly accelerated motion) 
 
2
0
2 2
1
0 ( )
2
1
(8 sin ) ( 9.81 m/s )
2
yy v t gt
t gt gα
= + −
= − =
 
At Point B: 21
0 (8 sin )
2B Bt gtα= − 
Simplifying and substituting for Bt 
 
1
0 8 sin
2 8 cos
d
gα
α
 
= −  
 
 
or 
64
sin 2d
g
α= (1) 
(a) When 0,h = the water can follow any physically possible trajectory. It then follows from 
 Eq. (1) that d is maximum when 2 90α = ° 
 or 45α = °  
 Then 
64
sin (2 45 )
9.81
d = × ° 
 or max 6.52 md = 

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