Throw a ball up in the air and it remains in the air for 8 sec. how fast did u throw it? what are the velocities after each second?

To solve this problem, we'll break it into parts using the principles of physics, particularly kinematics.

### Step 1: Determine the initial velocity
The ball is in the air for a total time of 8 seconds. Since this is a symmetric motion (up and down), it takes **4 seconds to reach the highest point**, and then another 4 seconds to return to the ground.

We can use the kinematic equation for vertical motion:
\[
v = u + at
\]
Where:
- \(v\) is the final velocity (which is 0 m/s at the highest point),
- \(u\) is the initial velocity (the velocity we want to find),
- \(a\) is the acceleration due to gravity, which is \(-9.8 \, \text{m/s}^2\) (negative because it acts downward),
- \(t\) is the time taken to reach the highest point (4 seconds).

At the highest point, the velocity is zero, so we can rearrange the equation to solve for \(u\):
\[
0 = u - 9.8 \times 4
\]
\[
u = 9.8 \times 4 = 39.2 \, \text{m/s}
\]
So, you threw the ball with an initial velocity of **39.2 m/s**.

### Step 2: Calculate the velocity after each second

To find the velocity after each second, we can again use the equation:
\[
v = u + at
\]
Where \(a = -9.8 \, \text{m/s}^2\).

#### 1 second:
\[
v_1 = 39.2 - 9.8 \times 1 = 39.2 - 9.8 = 29.4 \, \text{m/s}
\]

#### 2 seconds:
\[
v_2 = 39.2 - 9.8 \times 2 = 39.2 - 19.6 = 19.6 \, \text{m/s}
\]

#### 3 seconds:
\[
v_3 = 39.2 - 9.8 \times 3 = 39.2 - 29.4 = 9.8 \, \text{m/s}
\]

#### 4 seconds (highest point):
\[
v_4 = 39.2 - 9.8 \times 4 = 39.2 - 39.2 = 0 \, \text{m/s}
\]
At this point, the ball starts falling back down.

#### 5 seconds (falling down):
\[
v_5 = 0 - 9.8 \times 1 = -9.8 \, \text{m/s}
\]

#### 6 seconds:
\[
v_6 = 0 - 9.8 \times 2 = -19.6 \, \text{m/s}
\]

#### 7 seconds:
\[
v_7 = 0 - 9.8 \times 3 = -29.4 \, \text{m/s}
\]

#### 8 seconds (just before hitting the ground):
\[
v_8 = 0 - 9.8 \times 4 = -39.2 \, \text{m/s}
\]
So, the velocities after each second are:
1. **29.4 m/s**
2. **19.6 m/s**
3. **9.8 m/s**
4. **0 m/s** (highest point)
5. **-9.8 m/s**
6. **-19.6 m/s**
7. **-29.4 m/s**
8. **-39.2 m/s** (just before impact)

### Tip:
The velocity after each second decreases by \(9.8 \, \text{m/s}\), which is the acceleration due to gravity. This remains constant during free fall.

### Follow-up questions:
1. How does air resistance affect the time the ball stays in the air?
2. What would be the initial velocity if the ball stayed in the air for 10 seconds?
3. Can you calculate the maximum height the ball reaches?
4. What is the significance of negative velocity after the ball reaches the highest point?
5. How would the problem change on a different planet with a different gravitational acceleration?

Let me know if you'd like more details on any part of the solution!

Determine the initial velocity required to keep a ball in the air for 8 seconds and find the velocities after each second. This problem covers key concepts in kinematics, including projectile motion and acceleration due to gravity, with detailed step-by-step solutions. Suitable for Grades 9-12 students.

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