II. Two dimensional motion Projectile motion Initial horizontal and vertical velocity Viy= Vi [sine ] Vix Vi [ cos0 ]; = Final horizontal and vertical velocity Vfx = Vix =Vi [ cose ]; Vfy = Viy-g(t) 0= 45° will have a maximum horizontal displacement (X dirction) 5. A projectile with an initial velocity magnitude of 10 m/s, the initial velocity is making a 30° angle above the horizontal line. Find (a) the initial horizontal component of the initial velocity (b) the initial vertical component of the initial velocity, (c) how much time will the projectile take to fly up to the maximum height? Horizonal Uniform Circular Motion V=2π (r)/ (t), ac = V²/(r); Fe = mac 6. An 2 kg object is making a horizontal uniform circular motion. The radius of the circle is 2 meters and it makes 12 revolutions per minute. Find (a) it speed, (b) the centripetal acceleration magnitude of this object, (c) the centripetal force magnitude on the object.

College Physics
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ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter2: Motion In One Dimension
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Problem 66AP: Two students air on a balcony a distance h above the street. One student throws a ball vertically...
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Answer questions 5 and 6
## Two Dimensional Motion

### Projectile Motion

**Initial Horizontal and Vertical Velocity**
- \( V_{ix} = V_i \cdot \cos(\theta) \)
- \( V_{iy} = V_i \cdot \sin(\theta) \)

**Final Horizontal and Vertical Velocity**
- \( V_{fx} = V_{ix} = V_i \cdot \cos(\theta) \)
- \( V_{fy} = V_{iy} - g \cdot t \)

**Note:** \(\theta = 45^\circ\) will have a maximum horizontal displacement (X direction)

**Example Problem:**
5. A projectile with an initial velocity magnitude of 10 m/s, the initial velocity is making a \( 30^\circ \) angle above the horizontal line. Find:
   - (a) The initial horizontal component of the initial velocity.
   - (b) The initial vertical component of the initial velocity.
   - (c) How much time will the projectile take to fly up to the maximum height?

### Horizontal Uniform Circular Motion
- \( V = \frac{2\pi (r)}{t} \)
- \( a_c = \frac{V^2}{r} \)
- \( F_c = m \cdot a_c \)

**Example Problem:**
6. An object with a mass of 2 kg is making horizontal uniform circular motion. The radius of the circle is 2 meters and it makes 12 revolutions per minute. Find:
   - (a) Its speed.
   - (b) The centripetal acceleration magnitude of this object.
   - (c) The centripetal force magnitude on the object.

### Explanation of Concepts
**Projectile Motion:**
- **Horizontal and Vertical Components:** Any projectile’s motion can be analyzed by splitting its velocity into two components: horizontal (x-axis) and vertical (y-axis).
- **Initial Velocity Components:** Using trigonometry, the initial velocity \( V_i \) can be split into horizontal (\( V_{ix} \)) and vertical (\( V_{iy} \)) components.
- **Final Velocity:** The horizontal component remains constant in the absence of air resistance, whereas the vertical component changes due to gravitational acceleration (g).

**Horizontal Uniform Circular Motion:**
- **Speed (V):** The linear speed of an object in circular motion is given by the circumference of the circle (
Transcribed Image Text:## Two Dimensional Motion ### Projectile Motion **Initial Horizontal and Vertical Velocity** - \( V_{ix} = V_i \cdot \cos(\theta) \) - \( V_{iy} = V_i \cdot \sin(\theta) \) **Final Horizontal and Vertical Velocity** - \( V_{fx} = V_{ix} = V_i \cdot \cos(\theta) \) - \( V_{fy} = V_{iy} - g \cdot t \) **Note:** \(\theta = 45^\circ\) will have a maximum horizontal displacement (X direction) **Example Problem:** 5. A projectile with an initial velocity magnitude of 10 m/s, the initial velocity is making a \( 30^\circ \) angle above the horizontal line. Find: - (a) The initial horizontal component of the initial velocity. - (b) The initial vertical component of the initial velocity. - (c) How much time will the projectile take to fly up to the maximum height? ### Horizontal Uniform Circular Motion - \( V = \frac{2\pi (r)}{t} \) - \( a_c = \frac{V^2}{r} \) - \( F_c = m \cdot a_c \) **Example Problem:** 6. An object with a mass of 2 kg is making horizontal uniform circular motion. The radius of the circle is 2 meters and it makes 12 revolutions per minute. Find: - (a) Its speed. - (b) The centripetal acceleration magnitude of this object. - (c) The centripetal force magnitude on the object. ### Explanation of Concepts **Projectile Motion:** - **Horizontal and Vertical Components:** Any projectile’s motion can be analyzed by splitting its velocity into two components: horizontal (x-axis) and vertical (y-axis). - **Initial Velocity Components:** Using trigonometry, the initial velocity \( V_i \) can be split into horizontal (\( V_{ix} \)) and vertical (\( V_{iy} \)) components. - **Final Velocity:** The horizontal component remains constant in the absence of air resistance, whereas the vertical component changes due to gravitational acceleration (g). **Horizontal Uniform Circular Motion:** - **Speed (V):** The linear speed of an object in circular motion is given by the circumference of the circle (
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