SOLUTION:
(a) Using Equation 4-10, the maximum horizontal distance is
1 (5) V32.2 (2?
(b) Using Equation 4-11, the maximum horizontal velocity is
H^/gT
um<.» =
2d
1^32.2 (2)'
„
— --------------- = 2.0 reet per second .
Umax
2 (2)
(c) Using the relation L = T/gd to détermine the shallow-water
wavelength,
L = 5^32.2 (2? = 40.1 feet.
From (a) above, the maximum horizontal distance 2A is 3.2 feet
therefore the ratio 2A/L is
— = — = 0.08 .
L
40.1
(d) Using the relation C = /Jd (Equation 2-9) to détermine the
shallow-water wave speed
C = s/32.2 (2) = 8.0 feet per second .
From (b)
feet per
above the maximum horizontal velocity u^^,
second. Therefore the ratio
is
is 2.0
_ 0 25
C
8.0
********
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4-41