To determine the maximum height reached by the ball, we can use the following kinematic equation that relates the initial velocity, final velocity, acceleration due to gravity, and displacement (height):

$v^2 = u^2 + 2a s$

Where:

• $v$ is the final velocity at the maximum height (which is 0 m/s, since the ball momentarily stops before falling back down),
• $u$ is the initial velocity (30 m/s),
• $a$ is the acceleration due to gravity (-9.8 m/s², negative because it's acting downward),
• $s$ is the maximum height (what we're solving for).

Let's plug in the values:

$0^2 = (30)^2 + 2(-9.8)(s)$

Solving for $s$:

$0 = 900 - 19.6s$ $19.6s = 900$ $s = \frac{900}{19.6} \approx 45.92 \, \text{m}$

Thus, the closest value for the maximum height is approximately 46 meters.

Would you like more details or have any further questions?

Here are 5 related questions:

1. What is the total time of flight for the ball?
2. How long does it take the ball to reach the maximum height?
3. What would the velocity of the ball be after 2 seconds of flight?
4. If the ball is launched at an angle instead of vertically, how does that affect the maximum height?
5. How does air resistance affect the motion of the ball?

Tip: When solving projectile motion problems, always pay attention to whether you are working with vertical or horizontal components of motion separately.