A ball thrown vertically upwards from the ground level ,and hits the ground after 4 seconds. Calculate the maximum height it reached during its journey

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.

A ball thrown vertically upwards from the ground level, and hits the ground after 4 seconds. Calculate the maximum height it reached during its journey.

Discover how to calculate the maximum height reached by a ball thrown vertically upwards with a total air time of 4 seconds. This detailed solution covers key kinematic concepts, including the use of gravity and equations of motion. Perfect for high school students in Grades 9-12.

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