Here are the answers, please tell me if you have any problem.

1.
Initial velocity: 𝑣0 = 150 𝑚/𝑠𝑒𝑐
Elevation angle: 𝛼 = 60°
Initial height: ℎ = 0 𝑚
Gravitational acceleration: 𝑔 = 9.8 𝑚/𝑠 2
(a) The horizontal displacement: 𝑥 = 𝑣0 𝑡 cos 𝛼 = 150𝑡 cos 60° = 75𝑡
𝒙 = 𝟕𝟓𝒕
The vertical displacement:
1
1
𝑦 = ℎ + 𝑣0 𝑡 sin 𝛼 − 𝑔𝑡 2 = 0 + 150𝑡 sin 60° − × 9.8 × 𝑡 2
2
2
𝒚 = 𝟕𝟓√𝟑𝒕 − 𝟒. 𝟗𝒕𝟐
The particle landed when y =0
∴ 𝑦 = 75√3𝑡 − 4.9𝑡 2 = 0 ∴ 75√3 − 4.9𝑡 = 0 ∴ 𝑡 =

75√3
= 26.51 𝑠
4.9

Then 𝑥 = 75𝑡 = 75 × 26.51 = 1988.32 ≈ 1988 𝑚
Then, the point of landing: (𝑥, 𝑦) = (𝟏𝟗𝟖𝟖, 𝟎)
1

1

2

2

(b) The landing point is: (𝑥, 𝑦) = (𝑣0 𝑡 cos 𝛼 , 𝑣0 𝑡 sin 𝛼 − 𝑔𝑡 2 , 𝑣0𝑧 𝑡 + 𝑎𝑧 𝑡 2 )
Where: initial velocity in z-direction: 𝑣0𝑧 = 0 𝑚/𝑠
𝑎𝑧 is the lateral acceleration in z-direction.
So, the landing point when y =0 is:
1

1

2

2

(𝑥, 𝑦, 𝑧) = (𝑣0 𝑡 cos 𝛼 , 𝑣0 𝑡 sin 𝛼 − 𝑔𝑡 2 , − 𝑎𝑧 𝑡 2 )
1- 𝒗𝟎 = 𝟏𝟕𝟎 𝒎/𝒔 , 𝜶 = 𝟔𝟓°, 𝒂𝒛 = +𝟏 𝒎/𝒔𝟐 :
1
2 × 170 sin 65°
∴ 𝑦 = 170𝑡 sin 65° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 31.44 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (170 × 31.44 × cos 65° , 0, × 1 × (31.44)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟐𝟐𝟓𝟗, 𝟎, 𝟒𝟗𝟒)

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2- 𝒗𝟎 = 𝟏𝟕𝟎 𝒎/𝒔 , 𝜶 = 𝟔𝟓°, 𝒂𝒛 = −𝟏 𝒎/𝒔𝟐 :
1
2 × 170 sin 65°
∴ 𝑦 = 170𝑡 sin 65° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 31.44 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (170 × 31.44 × cos 65° , 0, × (−1) × (31.44)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟐𝟐𝟓𝟗, 𝟎, − 𝟒𝟗𝟒)
3- 𝒗𝟎 = 𝟏𝟕𝟎 𝒎/𝒔 , 𝜶 = 𝟓𝟓°, 𝒂𝒛 = +𝟏 𝒎/𝒔𝟐 :
1
2 × 170 sin 55°
∴ 𝑦 = 170𝑡 sin 55° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 28.42 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (170 × 28.42 × cos 65° , 0, × 1 × (28.42)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟐𝟎𝟒𝟐, 𝟎, 𝟒𝟎𝟒)
4- 𝒗𝟎 = 𝟏𝟕𝟎 𝒎/𝒔 , 𝜶 = 𝟓𝟓°, 𝒂𝒛 = −𝟏 𝒎/𝒔𝟐 :
1
2 × 170 sin 55°
∴ 𝑦 = 170𝑡 sin 55° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 28.42 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (170 × 28.42 × cos 65° , 0, × (−1) × (28.42)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟐𝟎𝟒𝟐, 𝟎, −𝟒𝟎𝟒)
5- 𝒗𝟎 = 𝟏𝟑𝟎 𝒎/𝒔 , 𝜶 = 𝟔𝟓°, 𝒂𝒛 = +𝟏 𝒎/𝒔𝟐 :
1
2 × 130 sin 65°
∴ 𝑦 = 130𝑡 sin 65° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 24.04 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (130 × 24.04 × cos 65° , 0, × 1 × (24.04)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟏𝟑𝟐𝟏, 𝟎, 𝟐𝟖𝟗)
𝒎

6- 𝒗𝟎 = 𝟏𝟑𝟎 , 𝜶 = 𝟔𝟓°, 𝒂𝒛 = −𝟏 𝒎/𝒔𝟐 :
𝒔

1
2 × 130 sin 65°
∴ 𝑦 = 130𝑡 sin 65° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 24.04 𝑠
2
9.8
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So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (130 × 24.04 × cos 65° , 0, × (−1) × (24.04)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟏𝟑𝟐𝟏, 𝟎, −𝟐𝟖𝟗)
7- 𝒗𝟎 = 𝟏𝟑𝟎 𝒎/𝒔 , 𝜶 = 𝟓𝟓°, 𝒂𝒛 = +𝟏 𝒎/𝒔𝟐 :
1
2 × 130 sin 55°
∴ 𝑦 = 130𝑡 sin 55° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 21.73 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (130 × 21.73 × cos 55° , 0, × 1 × (21.73)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟏𝟔𝟐𝟎, 𝟎, 𝟐𝟑𝟔)
8- 𝒗𝟎 = 𝟏𝟑𝟎 𝒎/𝒔 , 𝜶 = 𝟓𝟓°, 𝒂𝒛 = −𝟏 𝒎/𝒔𝟐 :
1
2 × 130 sin 55°
∴ 𝑦 = 130𝑡 sin 55° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 21.73 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (130 × 21.73 × cos 55° , 0, × (−1) × (21.73)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟏𝟔𝟐𝟎, 𝟎, −𝟐𝟑𝟔)
(c) The landing point in part (a) is: (1988, 0,0)
The farthest points from the center point (1988, 0,0) are: (1321, 0, −289)
and (1321, 0, 289) and they have equal distance from the center.
The radius of the circle is the distance between the points (1988, 0,0)
and(1321, 0, 289):
𝑟 = √(1321 − 1988)2 + (0 − 0)2 + (289 − 0)2 = 𝟕𝟐𝟕 𝒎

Page | 3

2.

x
𝐹𝑡ℎ
𝐹𝑡ℎ
y
𝐹𝑁

𝑊 cos 𝜃

𝜇𝑘 𝐹𝑁
𝑊 sin 𝜃

𝑊

85°

𝐹𝑁 = 𝑊 cos 𝜃 = 8 cos 85° = 0.6972 lb
- For 𝟎 < 𝒕 ≤ 𝟎. 𝟐
𝐹𝑡ℎ − 𝜇𝑘 𝐹𝑁 − 𝑊 sin 𝜃 =
∴𝑎=

𝑊
8
𝑎 ∴ 16 − 0.3 × 0.6972 − 8 sin 85° =
𝑎
𝑔
32.2

32.2
× (16 − 0.3 × 0.6972 − 8 sin 85°) = 31.48 𝑓𝑡/𝑠 2
8

To get the displacement:
1
1
𝑑1 = 𝑣0 𝑡 + 𝑎𝑡 2 = 0 × 0.2 + × 31.48 × 𝑡 2 = 𝟏𝟓. 𝟕𝟒 𝒕𝟐 𝒇𝒕
2
2
At t= 0.2 s, 𝑑1 = 15.74 × 0.22 = 0.6296 𝑓𝑡
The horizontal displacement: 𝑥1 = 𝑑1 𝑐𝑜𝑠𝜃 = 15.74 𝑡 2 cos 85° = 𝟏. 𝟑𝟕 𝒕𝟐 𝒇𝒕
The vertical displacement: 𝑦1 = 𝑑1 sin 𝜃 = 15.74 𝑡 2 sin 85° = 𝟏𝟓. 𝟔𝟖 𝒕𝟐 𝒇𝒕
- To get the final velocity at the first interval which is equal to initial velocity of the
seco...