Material is thrown from a conveyor at 60 ft/s at 43° above horizontal onto ground at the same level. Find the flight time, range and maximum height.

Start with the thinking

• Horizontal motion is constant velocity; vertical motion is constant acceleration.
• Time links the two components.

Step-by-step solution

1. Components

   $$v_{0}ₓ = 60cos 43^{\circ} = 43.88 ft/s, v_{0}ᵧ = 60sin 43^{\circ} = 40.92 ft/s$$

2. Flight time

   $$t = 2v_{0}ᵧ/g$$

3. Substituting

   $$t = 2(40.92)/32.2 = 2.542 s$$

4. Range — R = v₀ₓ·t = 43.88(2.542) = 111.5 ft

5. Peak height

   $$H = v_{0}ᵧ^{2}/(2g) = 26.00 ft$$

Why the other options are there

• R = 152.5 ft (full speed used horizontally)
• H = 52.0 ft (factor of 2 dropped)

Reference: FE Reference Handbook — Dynamics → Particle Curvilinear Motion

A car travels a curve of radius 140.0 m at 8 m/s while speeding up at 1.5 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Start with the thinking

• The tangential component changes speed; the normal component changes direction and always points to the centre.
• Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

1. Formula

   $$a_n = v^{2}/\rho$$

2. Substituting

   $$a_n = 8^{2}/140.0 = 0.457 m/s^{2}$$

3. Formula

   $$a = \sqrt(a_t^{2} + a_n^{2})$$

4. Substituting

   $$a = \sqrt(1.5^{2} + 0.457^{2}) = 1.568 m/s^{2}$$

5. Formula

   $$\omega = v/\rho$$

6. Substituting

   $$\omega = 8/140.0 = 0.0571 rad/s$$

Why the other options are there

• 1.96 m/s² (components added directly)
• 8,960 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Particle Curvilinear Motion

Material is thrown from a conveyor at 49 ft/s at 33° above horizontal onto ground at the same level. Find the flight time, range and maximum height.

Start with the thinking

• Horizontal motion is constant velocity; vertical motion is constant acceleration.
• Time links the two components.

Step-by-step solution

1. Components

   $$v_{0}ₓ = 49cos 33^{\circ} = 41.09 ft/s, v_{0}ᵧ = 49sin 33^{\circ} = 26.69 ft/s$$

2. Flight time

   $$t = 2v_{0}ᵧ/g$$

3. Substituting

   $$t = 2(26.69)/32.2 = 1.658 s$$

4. Range — R = v₀ₓ·t = 41.09(1.658) = 68.1 ft

5. Peak height

   $$H = v_{0}ᵧ^{2}/(2g) = 11.06 ft$$

Why the other options are there

• R = 81 ft (full speed used horizontally)
• H = 22.1 ft (factor of 2 dropped)

Reference: FE Reference Handbook — Dynamics → Particle Curvilinear Motion

A car travels a curve of radius 140.0 m at 32 m/s while speeding up at 2.5 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Start with the thinking

• The tangential component changes speed; the normal component changes direction and always points to the centre.
• Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

1. Formula

   $$a_n = v^{2}/\rho$$

2. Substituting

   $$a_n = 32^{2}/140.0 = 7.314 m/s^{2}$$

3. Formula

   $$a = \sqrt(a_t^{2} + a_n^{2})$$

4. Substituting

   $$a = \sqrt(2.5^{2} + 7.314^{2}) = 7.730 m/s^{2}$$

5. Formula

   $$\omega = v/\rho$$

6. Substituting

   $$\omega = 32/140.0 = 0.2286 rad/s$$

Why the other options are there

• 9.81 m/s² (components added directly)
• 143,360 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Particle Curvilinear Motion

Material is thrown from a conveyor at 86 ft/s at 49° above horizontal onto ground at the same level. Find the flight time, range and maximum height.

Start with the thinking

• Horizontal motion is constant velocity; vertical motion is constant acceleration.
• Time links the two components.

Step-by-step solution

1. Components

   $$v_{0}ₓ = 86cos 49^{\circ} = 56.42 ft/s, v_{0}ᵧ = 86sin 49^{\circ} = 64.91 ft/s$$

2. Flight time

   $$t = 2v_{0}ᵧ/g$$

3. Substituting

   $$t = 2(64.91)/32.2 = 4.031 s$$

4. Range — R = v₀ₓ·t = 56.42(4.031) = 227.5 ft

5. Peak height

   $$H = v_{0}ᵧ^{2}/(2g) = 65.41 ft$$

Why the other options are there

• R = 346.7 ft (full speed used horizontally)
• H = 130.8 ft (factor of 2 dropped)

Reference: FE Reference Handbook — Dynamics → Particle Curvilinear Motion

A car travels a curve of radius 125.0 m at 12 m/s while speeding up at 2.0 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Start with the thinking

• The tangential component changes speed; the normal component changes direction and always points to the centre.
• Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

1. Formula

   $$a_n = v^{2}/\rho$$

2. Substituting

   $$a_n = 12^{2}/125.0 = 1.152 m/s^{2}$$

3. Formula

   $$a = \sqrt(a_t^{2} + a_n^{2})$$

4. Substituting

   $$a = \sqrt(2.0^{2} + 1.152^{2}) = 2.308 m/s^{2}$$

5. Formula

   $$\omega = v/\rho$$

6. Substituting

   $$\omega = 12/125.0 = 0.0960 rad/s$$

Why the other options are there

• 3.15 m/s² (components added directly)
• 18,000 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Particle Curvilinear Motion

Material is thrown from a conveyor at 101 ft/s at 43° above horizontal onto ground at the same level. Find the flight time, range and maximum height.

Start with the thinking

• Horizontal motion is constant velocity; vertical motion is constant acceleration.
• Time links the two components.

Step-by-step solution

1. Components

   $$v_{0}ₓ = 101cos 43^{\circ} = 73.87 ft/s, v_{0}ᵧ = 101sin 43^{\circ} = 68.88 ft/s$$

2. Flight time

   $$t = 2v_{0}ᵧ/g$$

3. Substituting

   $$t = 2(68.88)/32.2 = 4.278 s$$

4. Range — R = v₀ₓ·t = 73.87(4.278) = 316.0 ft

5. Peak height

   $$H = v_{0}ᵧ^{2}/(2g) = 73.68 ft$$

Why the other options are there

• R = 432.1 ft (full speed used horizontally)
• H = 147.4 ft (factor of 2 dropped)

Reference: FE Reference Handbook — Dynamics → Particle Curvilinear Motion

A car travels a curve of radius 115.0 m at 21 m/s while speeding up at 1.5 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Start with the thinking

• The tangential component changes speed; the normal component changes direction and always points to the centre.
• Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

1. Formula

   $$a_n = v^{2}/\rho$$

2. Substituting

   $$a_n = 21^{2}/115.0 = 3.835 m/s^{2}$$

3. Formula

   $$a = \sqrt(a_t^{2} + a_n^{2})$$

4. Substituting

   $$a = \sqrt(1.5^{2} + 3.835^{2}) = 4.118 m/s^{2}$$

5. Formula

   $$\omega = v/\rho$$

6. Substituting

   $$\omega = 21/115.0 = 0.1826 rad/s$$

Why the other options are there

• 5.33 m/s² (components added directly)
• 50,715 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Particle Curvilinear Motion

Material is thrown from a conveyor at 118 ft/s at 57° above horizontal onto ground at the same level. Find the flight time, range and maximum height.

Start with the thinking

• Horizontal motion is constant velocity; vertical motion is constant acceleration.
• Time links the two components.

Step-by-step solution

1. Components

   $$v_{0}ₓ = 118cos 57^{\circ} = 64.27 ft/s, v_{0}ᵧ = 118sin 57^{\circ} = 98.96 ft/s$$

2. Flight time

   $$t = 2v_{0}ᵧ/g$$

3. Substituting

   $$t = 2(98.96)/32.2 = 6.147 s$$

4. Range — R = v₀ₓ·t = 64.27(6.147) = 395.0 ft

5. Peak height

   $$H = v_{0}ᵧ^{2}/(2g) = 152.1 ft$$

Why the other options are there

• R = 725.3 ft (full speed used horizontally)
• H = 304.2 ft (factor of 2 dropped)

Reference: FE Reference Handbook — Dynamics → Particle Curvilinear Motion

A car travels a curve of radius 85 m at 34 m/s while speeding up at 1.0 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Start with the thinking

• The tangential component changes speed; the normal component changes direction and always points to the centre.
• Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

1. Formula

   $$a_n = v^{2}/\rho$$

2. Substituting

   $$a_n = 34^{2}/85 = 13.600 m/s^{2}$$

3. Formula

   $$a = \sqrt(a_t^{2} + a_n^{2})$$

4. Substituting

   $$a = \sqrt(1.0^{2} + 13.600^{2}) = 13.637 m/s^{2}$$

5. Formula

   $$\omega = v/\rho$$

6. Substituting

   $$\omega = 34/85 = 0.4000 rad/s$$

Why the other options are there

• 14.60 m/s² (components added directly)
• 98,260 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Particle Curvilinear Motion