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Let's tackle each of the three problems step-by-step.

### 1. Problem 1
- **Given**:
  - Initial velocity, \( v_0 = 6.0 \, \text{m/s} \)
  - Initial height, \( y_0 = 2.0 \, \text{m} \)
  - Acceleration, \( a = -9.8 \, \text{m/s}^2 \) (due to gravity, acting downwards)
  - Final height when it hits the ground, \( y = 0 \, \text{m} \)

#### To find: 
**(a)** Final velocity just before it strikes the ground, \( v \).  
**(b)** Time of flight, \( t \).

##### Part (a) – Finding \( v \):
We can use the kinematic equation:
\[
v^2 = v_0^2 + 2a(y - y_0)
\]
Substituting the values:
\[
v^2 = (6.0)^2 + 2(-9.8)(0 - 2.0)
\]
\[
v^2 = 36 - 39.2
\]
\[
v^2 = -3.2
\]
\[
v = \sqrt{39.2} \approx 6.26 \, \text{m/s}
\]
(We take the negative value of \( v \) since the ball is moving downward when it strikes the floor.)

##### Part (b) – Finding \( t \):
Now, use the following equation to find \( t \):
\[
y = y_0 + v_0 t + \frac{1}{2} a t^2
\]
Since \( y = 0 \):
\[
0 = 2.0 + 6.0 t - 4.9 t^2
\]
This is a quadratic equation:
\[
4.9 t^2 - 6.0 t - 2.0 = 0
\]
Solving this quadratic using the formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4.9 \), \( b = -6.0 \), and \( c = -2.0 \):
\[
t = \frac{-(-6.0) \pm \sqrt{(-6.0)^2 - 4(4.9)(-2.0)}}{2(4.9)}
\]
\[
t = \frac{6.0 \pm \sqrt{36 + 39.2}}{9.8}
\]
\[
t = \frac{6.0 \pm \sqrt{75.2}}{9.8}
\]
\[
t = \frac{6.0 \pm 8.67}{9.8}
\]
We get two solutions for \( t \):
\[
t_1 = \frac{6.0 + 8.67}{9.8} = 1.5 \, \text{seconds}
\]
\[
t_2 = \frac{6.0 - 8.67}{9.8} \approx -0.27 \, \text{seconds} \quad (\text{not physically possible})
\]
Thus, the time in the air is \( t = 1.50 \, \text{seconds} \).

---

### 2. Problem 2
**Given**: Same values from Problem 1.

#### To find: 
The **maximum height** the ball will reach.

At maximum height, the velocity \( v = 0 \). Using the kinematic equation:
\[
v^2 = v_0^2 + 2a(y - y_0)
\]
Substitute \( v = 0 \) to solve for \( y \):
\[
0 = (6.0)^2 + 2(-9.8)(y - 2.0)
\]
\[
0 = 36 - 19.6(y - 2.0)
\]
\[
19.6(y - 2.0) = 36
\]
\[
y - 2.0 = \frac{36}{19.6} \approx 1.84
\]
\[
y = 1.84 + 2.0 = 3.84 \, \text{meters}
\]
So, the maximum height above the ground is **3.84 m**.

---

### 3. Problem 3
**Given**:
- Initial velocity, \( v_0 = 7.5 \, \text{m/s} \)
- Initial height, \( y_0 = 0.80 \, \text{m} \)
- Acceleration, \( a = -9.8 \, \text{m/s}^2 \)
- Final velocity at the maximum height, \( v = 0 \)

#### To find: 
The **maximum height** the ball will reach.

Using the same kinematic equation:
\[
v^2 = v_0^2 + 2a(y - y_0)
\]
Since \( v = 0 \), solve for \( y \):
\[
0 = (7.5)^2 + 2(-9.8)(y - 0.80)
\]
\[
0 = 56.25 - 19.6(y - 0.80)
\]
\[
19.6(y - 0.80) = 56.25
\]
\[
y - 0.80 = \frac{56.25}{19.6} \approx 2.87
\]
\[
y = 2.87 + 0.80 = 3.67 \, \text{meters}
\]
Thus, the maximum height reached is **3.67 meters**.

---

Do you want further details on any of the steps or clarifications?

Here are 5 related questions to practice:
1. How would the time of flight change if the initial velocity were doubled?
2. If air resistance were considered, how would that affect the final velocity?
3. What would happen if the ball was launched from a higher altitude, like 5.0 m?
4. Can you calculate the time taken to reach the maximum height in each scenario?
5. How would different gravity on another planet affect the maximum height?

**Tip**: When using the quadratic formula, ensure you check both solutions and choose the physically meaningful one (positive time).

A volleyball is hit straight up into the air with an initial velocity of 6.0 m/s. If the volleyball starts from 2.0 m above the floor, how long will it be in the air before it strikes the floor? (find v first) [1.50s]

Learn how to calculate the time of flight and the maximum height of a volleyball hit straight up into the air with an initial velocity. This problem covers key concepts in kinematics and projectile motion. Suitable for students in Grades 9-12.

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