When a ball is thrown vertically upwards it goes through a distance of 19.6 m. Find the initial velocity of the ball and the time taken by it to rise to the high point (Acceleration due to gravity g=9.8 m/s^(2))

Answer

To solve the problem, we will use the equations of motion under uniform acceleration. The ball is thrown vertically upwards, and we know the distance it travels (19.6 m) and the acceleration due to gravity (g = 9.8 m/s²).

### Step-by-Step Solution:

1. **Identify the Variables**:
- Distance (h) = 19.6 m (the height the ball reaches)
- Final velocity (v) at the highest point = 0 m/s (the ball stops rising at the highest point)
- Acceleration due to gravity (g) = -9.8 m/s² (negative because it acts downward)

2. **Use the Third Equation of Motion**:
The third equation of motion states:
\[ v^2 = u^2 + 2a s \]
where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration
- \( s \) = distance traveled

Since the final velocity \( v = 0 \) m/s at the highest point, we can rearrange the equation to solve for \( u \):
\[ 0 = u^2 - 2gh \]
Rearranging gives:
\[ u^2 = 2gh \]
Taking the square root:
\[ u = \sqrt{2gh} \]

3. **Substitute the Values**:
Now, substitute \( g = 9.8 \, \text{m/s}^2 \) and \( h = 19.6 \, \text{m} \):
\[ u = \sqrt{2 \times 9.8 \times 19.6} \]
Calculate:
\[ u = \sqrt{2 \times 9.8 \times 19.6} = \sqrt{384.16} = 19.6 \, \text{m/s} \]

4. **Calculate the Time Taken to Reach Maximum Height**:
We can use the formula:
\[ v = u - gt \]
At the maximum height, \( v = 0 \), so:
\[ 0 = u - gt \]
Rearranging gives:
\[ gt = u \]
Thus:
\[ t = \frac{u}{g} \]
Substitute \( u = 19.6 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \):
\[ t = \frac{19.6}{9.8} = 2 \, \text{s} \]

### Final Answers:
- Initial velocity of the ball, \( u = 19.6 \, \text{m/s} \)
- Time taken to reach the maximum height, \( t = 2 \, \text{s} \)