Webreview cp physics ch 3 practice test (holt)

ID: A
23. 1.09 m at 24° north of east
Given
d 1
= 2.00 m
d 2
= 2.50 m
θ 1
= 10.0° east of south
θ 2
= 75.0° north of east
Solution
Δx 1
= d 1
sinθ 1
= (2.00)(sin10.0°) = 0.347 m
Δy 1
= d 1
cosθ 1
= (2.00 m)(cos10.0°) = 1.97 m
Δx 2
= d 2
cosθ 2
= (2.50 m)(cos75.0°) = 0.647 m
Δy 2
= d 2
sinθ 2
= (2.50 m)(sin75.0°) = 2.41 m
Δx tot
= Δx 1
+ Δx 2
= 0.347 m + 0.647 m = 0.994 m
Δy tot
= Δy 1
+ Δy 2
= −1.97 m + 2.41 m = 0.44 m
d = (Δx tot
) 2 + (Δy tot
) 2 = (0.994 m) 2 + (0.44 m) 2 = 1.09 m
Ê 0.44 m ˆ
θ = tan −1
Ë
Á 0.994 m ¯
˜ = 24°
d = 1.09 m at 24° north of east
24. 120 m
Given
v i
= 12 m/s at 30.0° above the horizontal
Δt = 5.60 s
g = 9.81 m/s 2
Solution
v i, y
= v i
sinθ = (12 m/s)(sin30.0°) = 6.0 m/s
Δy = v i, y
Δt + 1 2 a y (Δt) 2 = (6.0 m/s)(5.60 s) + 1 2 (−9.81 m/s 2 )(5.60 s) 2
Δy = 34 m – 154 m = − 120 m
h = 120 m
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