## Description

I have in homework on Mechanics Fundamentals and I need help for this with it is asking for, no exert work.

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## Explanation & Answer

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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)

Page | 1

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

Page | 2

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...