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

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 491
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R1–17. A ball is launched from point A at an angle of 30°.
Determine the maximum and minimum speed v A it can
have so that it lands in the container.
A
v A
30
1 m
B
C
0.25 m
2.5 m
4 m
Min. speed:
a : + b s = s 0 + v 0 t
2.5 = 0 + v A cos 30°t
A + c B s = s 0 + v 0 t + 1 2 a c t 2
0.25 = 1 + v A sin 30°t - 1 2 (9.81)t2
Solving
t = 0.669 s
v A = Av A B min = 4.32 m>s
Ans.
Max. speed:
a : + b s = s 0 + v 0 t
4 = 0 + v A cos 30°t
A + c B s = s 0 + v 0 t + 1 2 a c t 2
0.25 = 1 + v A sin 30° t - 1 (9.81) t2
2
Solving:
t = 0.790 s
v A = (v A ) max = 5.85 m>s
Ans.
491