91962_01_s12-p0001-0176
6/8/09
8:35 AM
Page 83
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12–102. A golf ball is struck with a velocity of 80 ft>s as
shown. Determine the distance d to where it will land.
Horizontal Motion: The horizontal component of velocity is (y0)x = 80 cos 55°
= 45.89 ft>s.The initial and final horizontal positions are (s0)x = 0 and sx = d cos 10°,
respectively.
A
+
:
B
vA ⫽ 80 ft/s
B
A
45⬚
10⬚
d
sx = (s0)x + (y0)x t
d cos 10° = 0 + 45.89t
[1]
Vertical Motion: The vertical component of initial velocity is (y0)y = 80 sin 55°
= 65.53 ft>s. The initial and final vertical positions are (s0)y = 0 and sy = d sin 10°,
respectively.
(+ c )
1
(a ) t2
2 cy
1
d sin 10° = 0 + 65.53t + (-32.2)t2
2
sy = (s0)y + (y0)y t +
[2]
Solving Eqs. [1] and [2] yields
d = 166 ft
Ans.
t = 3.568 s
12–103. The football is to be kicked over the goalpost,
which is 15 ft high. If its initial speed is vA = 80 ft>s,
determine if it makes it over the goalpost, and if so, by how
much, h.
h
vA ⫽ 80 ft/s
B
15 ft
60⬚
Horizontal Motion: The horizontal component of velocity is (y0)x = 80 cos 60°
= 40.0 ft>s. The initial and final horizontal positions are (s0)x = 0 and sx = 25 ft,
respectively.
+ B
A:
sx = (s0)x + (y0)x t
25 = 0 + 40.0t
t = 0.625 s
Vertical Motion: The vertical component of initial velocity is (y0)y = 80 sin 60°
= 69.28 ft>s. The initial and final vertical positions are (s0)y = 0 and sy = H,
respectively.
A+cB
sy = (s0)y + (y0)y t +
H = 0 + 69.28(0.625) +
1
(a ) t2
2 cy
1
(-32.2) A 0.6252 B
2
H = 37.01 ft
Since H 7 15 ft , the football is kicked over the goalpost.
Ans.
h = H - 15 = 37.01 - 15 = 22.0 ft
Ans.
83
25 ft
45⬚
30 ft
x
y