https://file.aiquickdraw.com/mathbot/file/17272995557884kz3bou1.png

The problem gives the following information:

- Initial velocity (\(v_i\)) = 30 m/s
- Final velocity (\(v_f\)) at the maximum height = 0 m/s (since the ball comes to rest momentarily before falling back down)
- Acceleration (\(a\)) = -9.8 m/s² (the acceleration due to gravity is negative because it opposes the upward motion)
- We are solving for the displacement (\(d\)) or height.

The equation provided is:

\[
v_f^2 = v_i^2 + 2ad
\]

We can rearrange the equation to solve for \(d\) (the maximum height):

\[
d = \frac{v_f^2 - v_i^2}{2a}
\]

Substitute the known values:

\[
d = \frac{0^2 - 30^2}{2 \times (-9.8)}
\]
\[
d = \frac{-900}{-19.6}
\]
\[
d \approx 45.92 \text{ meters}
\]

Thus, the closest answer is **46 m (Option B)**.

Would you like more details on the solution or further clarification?

Here are 5 related questions you might find useful:
1. What is the significance of the negative acceleration in projectile motion?
2. How would the maximum height change if the initial velocity was doubled?
3. How does air resistance affect the maximum height reached in reality?
4. What is the time taken to reach the maximum height in this problem?
5. How can we calculate the total time the ball is in the air?

**Tip**: In projectile motion, always remember that the velocity at the highest point is zero, which simplifies calculations for maximum height.

A ball is projected vertically upward from the ground with an initial velocity of 30 m/s. Which of the following is closest to the maximum height reached by the ball?

Calculate the maximum height of a ball projected vertically upward with an initial velocity of 30 m/s using the kinematic equation v_f^2 = v_i^2 + 2ad. This problem applies core kinematic concepts and is ideal for high school students studying projectile motion.

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