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8. A negative discriminant indicates that the apex of the movement (the farthest displacement
in the direction of the initial velocity) is not large enough to reach the desired displacement
∆x. Thus, there is no value of t for which the sought-after displacement will be reached.
If the discriminant is zero, then there is exactly one solution to Equation (11.16), and the
displacement is equal to the maximum displacement at the apex.
Note that if the acceleration and the displacement have the same sign, then the discriminant
can never be positive, and there will always be two solutions except in the trivial case where
all the values are zero.
(a) v 0 = 150[cos 40 o , sin 40 o ] ft/s ≈ [114.9, 96.4] ft/s
(b) t = −(v 0 ) y /a y = −(96.4 ft/s)/(−32.0 ft/s 2 ) = 3.01 s
9.
0 ft/s 2
−32.0 ft/s 2
0 ft
10 ft
114.9 ft/s
96.4 ft/s
t + 2
t 2
(c) p(t) =
+
0 ft/s 2
−32.0 ft/s 2
0 ft
10 ft
114.9 ft/s
96.4 ft/s
(3.01 s) + 2
(3.01 s) 2
p(3.01 s) =
+
0 ft
10 ft
345.8 ft
290.2 ft
0 ft
−145.0 ft
345.8 ft
155.2 ft
=
+
+
=
(d) It's twice the time to reach the apex, 2(3.01 s) = 6.02 s.
(e) x(t) = (0 ft) + (114.9 ft/s)t + (1/2)(0 ft/s 2 )t 2
x(6.02 s) = (114.9 ft/s)(6.02 s) = 691.7 ft
10. ∆p = v 0 t + (1/2)at 2
∆pa = (v 0 t + (1/2)at 2 )a
∆pa = (v 0 a)t + (1/2)(aa)t 2
0 = (aa/2)t 2 + (v 0 a)t−∆pa
−(v 0 a)±
(v 0 a) 2 −4(aa/2)(∆pa)
2(aa/2)
t =
(v 0 a) 2 −2(aa)(∆pa)
aa
11. Expanding the Taylor series for e ix :
−(v 0 a)±
t =
e ix = 1 + ix + (ix) 2
2!
+ (ix) 3
3!
+ (ix) 4
4!
+ (ix) 5
5!
+ (ix) 6
6!
+ (ix) 7
7!
+ (ix) 8
8!
+
Substituting the powers of i (i 2 = −1, i 3 = −i, i 4 = 1, etc.):
e ix = 1 + ix− x 2
2!
ix 3
3!
+ x 4
4!
+ ix 5
5!
x 6
6!
ix 7
7!
+ x 8
8!
+
Now we separate the real and imaginary terms:
x− x 3
3!
1− x 2
2!
+ x 4
4!
x 6
6!
+ x 8
8!
+ x 5
5!
x 7
7!
e ix =
+ i
+
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