From the top of a tower of height 50m, a ball is thrown vertically upwards with a certain velocity. It hits the ground 10 s after it is thrown up. How much time does it take to cover a distance AB where A and B are two points 20m and 40m below the edge of the tower ? `(g= 10ms^(-2))`

Options

A. 2.0 s
B. 1.0 s
C. 0.5 s
D. 0.4 s

Answer

To solve the problem, we need to find the time it takes for the ball to travel between two points, A and B, which are located 20 m and 40 m below the top of a 50 m tower, respectively. The ball is thrown upwards and hits the ground after 10 seconds.

### Step-by-Step Solution:

1. **Identify the total height and the distances to points A and B**:
- The height of the tower is 50 m.
- Point A is 20 m below the top of the tower, so its height from the ground is \(50 - 20 = 30\) m.
- Point B is 40 m below the top of the tower, so its height from the ground is \(50 - 40 = 10\) m.

2. **Use the second equation of motion**:
The second equation of motion is given by:
\[ h = ut + \frac{1}{2} a t^2 \]
where \(h\) is the displacement, \(u\) is the initial velocity, \(a\) is the acceleration (which is \(-g\) or \(-10 \, \text{m/s}^2\) in this case), and \(t\) is the time.

3. **Find the initial velocity**:
Since the ball hits the ground after 10 seconds, we can use the total displacement from the top of the tower to the ground:
\[ -50 = u(10) - \frac{1}{2} (10)(10^2) \]
Simplifying this gives:
\[ -50 = 10u - 500 \]
\[ 10u = 450 \implies u = 45 \, \text{m/s} \]

4. **Calculate time to reach point A (30 m from the ground)**:
For point A, the displacement is:
\[ h = -20 \, \text{m} \quad (\text{since it is 20 m below the top}) \]
Using the second equation of motion:
\[ -20 = 45t_1 - \frac{1}{2} (10)t_1^2 \]
Rearranging gives:
\[ 5t_1^2 - 45t_1 - 20 = 0 \]
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ t_1 = \frac{45 \pm \sqrt{(-45)^2 - 4 \cdot 5 \cdot (-20)}}{2 \cdot 5} \]
\[ t_1 = \frac{45 \pm \sqrt{2025 + 400}}{10} \]
\[ t_1 = \frac{45 \pm \sqrt{2425}}{10} \]
Approximating \(\sqrt{2425} \approx 49.25\):
\[ t_1 = \frac{45 + 49.25}{10} \approx 9.42 \, \text{s} \quad (\text{taking the positive root}) \]

5. **Calculate time to reach point B (10 m from the ground)**:
For point B, the displacement is:
\[ h = -40 \, \text{m} \]
Using the second equation of motion:
\[ -40 = 45t_2 - \frac{1}{2} (10)t_2^2 \]
Rearranging gives:
\[ 5t_2^2 - 45t_2 - 40 = 0 \]
Using the quadratic formula:
\[ t_2 = \frac{45 \pm \sqrt{(-45)^2 - 4 \cdot 5 \cdot (-40)}}{2 \cdot 5} \]
\[ t_2 = \frac{45 \pm \sqrt{2025 + 800}}{10} \]
\[ t_2 = \frac{45 \pm \sqrt{2825}}{10} \]
Approximating \(\sqrt{2825} \approx 53.13\):
\[ t_2 = \frac{45 + 53.13}{10} \approx 9.81 \, \text{s} \]

6. **Calculate the time taken to travel from A to B**:
The time taken to travel from point A to point B is:
\[ t_{AB} = t_2 - t_1 = 9.81 - 9.42 = 0.39 \, \text{s} \]

### Final Answer:
The time taken to cover the distance from point A to point B is approximately **0.39 seconds**.