6. A projectile was launched vertically upward with an initial speed of 30.0 m/s from the ground. Find the time it takes to get to the maximum height. Find the maximum height of the projectile. b. Find its position 2 seconds later. When is the speed of the projectile V = -30m/s

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### Problem 6: Vertical Projectile Motion

A projectile was launched vertically upward with an initial speed of 30.0 m/s from the ground. The following queries should be addressed:

#### a. Find the maximum height of the projectile.

#### b. Find its position 2 seconds later.

#### c. Determine when the speed of the projectile is \( V = -30 \, \mathrm{m/s} \).

#### Relevant Equations:

1. \( v_f - v_i = -gt \)
2. \( y = v_i t - \frac{1}{2} g t^2 \)

Where:
- \( v_f \) is the final velocity
- \( v_i \) is the initial velocity
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s²)
- \( t \) is the time
- \( y \) is the displacement

#### Calculations:

**a. Maximum Height**
- At maximum height, the final velocity \( v_f = 0 \).

**b. Position 2 Seconds Later**
- Use the second equation to find the displacement at \( t = 2 \) seconds.

**c. Speed of the Projectile When \( V = -30 \, \mathrm{m/s} \)**
- Use the first equation to solve for \( t \).

These steps provide a comprehensive method to analyze the motion of a projectile launched vertically.
Transcribed Image Text:### Problem 6: Vertical Projectile Motion A projectile was launched vertically upward with an initial speed of 30.0 m/s from the ground. The following queries should be addressed: #### a. Find the maximum height of the projectile. #### b. Find its position 2 seconds later. #### c. Determine when the speed of the projectile is \( V = -30 \, \mathrm{m/s} \). #### Relevant Equations: 1. \( v_f - v_i = -gt \) 2. \( y = v_i t - \frac{1}{2} g t^2 \) Where: - \( v_f \) is the final velocity - \( v_i \) is the initial velocity - \( g \) is the acceleration due to gravity (approximately 9.8 m/s²) - \( t \) is the time - \( y \) is the displacement #### Calculations: **a. Maximum Height** - At maximum height, the final velocity \( v_f = 0 \). **b. Position 2 Seconds Later** - Use the second equation to find the displacement at \( t = 2 \) seconds. **c. Speed of the Projectile When \( V = -30 \, \mathrm{m/s} \)** - Use the first equation to solve for \( t \). These steps provide a comprehensive method to analyze the motion of a projectile launched vertically.
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