To solve this problem, we'll use the principles of kinematics. When a ball is thrown vertically upward, it will rise to a maximum height, momentarily stop, and then fall back to the ground. The time it takes to reach the maximum height is half the total time in the air.

Step-by-Step Solution:

1. Total time in the air (T):
   The ball hits the ground after 4 seconds. Since the time taken to go up is equal to the time taken to come down, the time to reach the maximum height is: $T_{\text{up}} = \frac{T}{2} = \frac{4 \text{ s}}{2} = 2 \text{ s}$

2. Acceleration due to gravity (g):
   We know that $g = 9.8 \, \text{m/s}^2$ (acting downward).

3. Initial velocity (u):
   We can calculate the initial velocity using the formula: $v = u - g \cdot t$ At the maximum height, the final velocity $v = 0$, so: $0 = u - g \cdot T_{\text{up}}$ Solving for $u$: $u = g \cdot T_{\text{up}} = 9.8 \, \text{m/s}^2 \times 2 \, \text{s} = 19.6 \, \text{m/s}$

4. Maximum height (H):
   The maximum height can be calculated using the formula: $H = u \cdot t - \frac{1}{2} g \cdot t^2$ Substitute the values $u = 19.6 \, \text{m/s}$, $t = 2 \, \text{s}$, and $g = 9.8 \, \text{m/s}^2$: $H = 19.6 \, \text{m/s} \times 2 \, \text{s} - \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (2 \, \text{s})^2$ Simplifying: $H = 39.2 \, \text{m} - \frac{1}{2} \times 9.8 \times 4$ $H = 39.2 \, \text{m} - 19.6 \, \text{m} = 19.6 \, \text{m}$

Final Answer:

The maximum height reached by the ball is 19.6 meters.

Would you like more details on this solution or have any questions?

Here are 5 related questions to explore further:

1. How would the maximum height change if the ball took 6 seconds to hit the ground?
2. What would be the initial velocity if the total time in the air was 5 seconds?
3. Calculate the maximum height if the acceleration due to gravity was different, say 10 m/s².
4. Determine the time it would take to reach half the maximum height.
5. How would air resistance affect the maximum height?

Tip: When solving kinematics problems, clearly separate the time spent rising and falling to simplify calculations.