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Do not solve this for me but instead show me how to solve a very similar problem

To solve a problem similar to the one you shared, where an object is thrown upward with an initial velocity and you are asked to find both the maximum height and the total time it stays in the air, we use the principles of kinematics under constant acceleration (due to gravity).

### Let’s break down the process:

### Given Information (Hypothetical Similar Problem):
- **Initial velocity (v₀):** \( v_0 = 22 \, \text{m/s} \)
- **Final velocity at the peak (v):** At the highest point, the velocity is zero, so \( v = 0 \, \text{m/s} \).
- **Acceleration due to gravity (g):** \( g = 9.8 \, \text{m/s}^2 \) (downward, hence we'll use it as negative: \( -9.8 \, \text{m/s}^2 \)).
- **Total displacement to find (h for height):** unknown
- **Time (t):** unknown for total air time

### Part (a): How high does it go?
To find the maximum height, we use the following kinematic equation:

\[
v^2 = v_0^2 + 2a d
\]
Here:
- \( v \) is the final velocity at the peak (which is 0 m/s).
- \( v_0 \) is the initial velocity (22 m/s in this case).
- \( a \) is the acceleration due to gravity, \( a = -9.8 \, \text{m/s}^2 \).
- \( d \) is the maximum height we are solving for (h).

**Steps:**
1. Plug in the values for \( v \), \( v_0 \), and \( a \).
2. Solve for \( d \), the maximum height.

### Part (b): How long is it in the air?
To find the total time, we calculate the time to reach the peak and then double it (since the time to go up equals the time to come down).

To find the time to reach the peak, we use the equation:

\[
v = v_0 + a t
\]
Here:
- \( v \) is the final velocity at the peak (0 m/s),
- \( v_0 \) is the initial velocity (22 m/s),
- \( a \) is the acceleration due to gravity (\(-9.8 \, \text{m/s}^2\)),
- \( t \) is the time to reach the maximum height.

**Steps:**
1. Plug in the values and solve for \( t \).
2. Double the time found (since the time to go up is the same as the time to come down).

### Now that you understand the method, you can apply these steps to your specific problem by plugging in the actual values.

Do you have any questions about this approach or need further details on any step?

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Here are 5 relative questions for further practice:

1. What would happen to the maximum height if the initial speed was doubled?
2. How would the total time in the air change if the initial velocity was half of the given value?
3. What is the impact of air resistance on the calculations we performed?
4. Can we use energy conservation principles to solve for the maximum height instead of kinematic equations?
5. How would the motion differ if the object was thrown at an angle instead of vertically?

**Tip:** Always remember to carefully consider the direction of the acceleration due to gravity—it is negative when dealing with upward motion.

A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) How high does it go? (b) How long is it in the air?

Learn how to calculate the maximum height and total air time for a baseball hit with an initial velocity of 22 m/s using kinematic equations. This physics problem covers projectile motion and is suitable for high school students (Grades 9-12).

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