a ball is thrown straight up in the air with a speed of 16 meters a second

When a ball is thrown straight up into the air with an initial speed of 16 meters per second, we can analyze its motion using basic kinematic equations. Since the only force acting on the ball (assuming air resistance is negligible) is gravity, we can describe its motion in terms of the following parameters:

1. **Initial velocity (v₀):** 16 m/s (upward)
2. **Acceleration due to gravity (g):** 9.8 m/s² (downward)
3. **Maximum height (h_max):** The highest point the ball reaches, where its velocity becomes zero.
4. **Time to reach the maximum height (t_up):** The time taken for the ball to stop moving upward.
5. **Total time of flight (t_total):** The total time the ball is in the air, including both upward and downward motion.

### Calculating the Time to Reach Maximum Height
At the maximum height, the velocity of the ball becomes zero. Using the following equation of motion:
\[
v = v_0 - g t
\]
Where:
- \(v\) is the final velocity (0 m/s at the top),
- \(v_0\) is the initial velocity (16 m/s),
- \(g\) is the acceleration due to gravity (9.8 m/s²),
- \(t\) is the time to reach the maximum height.

Setting \(v = 0\), we can solve for \(t\):
\[
0 = 16 - 9.8 t \quad \Rightarrow \quad t = \frac{16}{9.8} \approx 1.63 \text{ seconds}.
\]

### Maximum Height Reached
To find the maximum height, we can use the equation:
\[
h = v_0 t - \frac{1}{2} g t^2
\]
Substituting \(v_0 = 16 \, \text{m/s}\), \(g = 9.8 \, \text{m/s}^2\), and \(t = 1.63 \, \text{seconds}\), we get:
\[
h = 16 \times 1.63 - \frac{1}{2} \times 9.8 \times (1.63)^2 \approx 13.06 \, \text{meters}.
\]

### Total Time of Flight
Since the ball takes 1.63 seconds to reach its maximum height, the time to fall back down will be the same. Thus, the total time of flight is:
\[
t_{\text{total}} = 2 \times 1.63 \approx 3.26 \, \text{seconds}.
\]

### Summary of Results:
1. Time to reach the maximum height: \( t \approx 1.63 \, \text{seconds} \)
2. Maximum height reached: \( h_{\text{max}} \approx 13.06 \, \text{meters} \)
3. Total time of flight: \( t_{\text{total}} \approx 3.26 \, \text{seconds} \)

Would you like a more detailed explanation of any step?

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Here are 5 related questions:
1. How would air resistance affect the ball’s motion?
2. What would happen if the initial velocity was doubled?
3. How does the maximum height depend on the initial velocity?
4. What is the velocity of the ball when it returns to the ground?
5. Can we calculate the displacement of the ball after 1 second?

**Tip:** Always keep track of signs (+/-) when solving kinematic equations to distinguish between upward and downward motion.

A ball is thrown straight up in the air with a speed of 16 meters a second.

Explore how to analyze the motion of a ball thrown straight up into the air with an initial speed of 16 m/s. Learn to calculate the time to reach maximum height, the maximum height, and the total time of flight using kinematic equations. Suitable for Grades 9-12.

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