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 step-by-step, we will use the equations of motion under gravity. The ball is thrown upwards from a height of 50 m and hits the ground after 10 seconds. We need to find the time taken to cover the distance between points A and B, which are located 20 m and 40 m below the edge of the tower, respectively.

### Step 1: Determine the initial velocity (u) of the ball
We know the following:
- Height of the tower (s) = 50 m (downward)
- Time taken to hit the ground (t) = 10 s
- Acceleration due to gravity (g) = 10 m/s² (downward)

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ -50 = u(10) + \frac{1}{2}(-10)(10^2) \]
\[ -50 = 10u - 500 \]
Rearranging gives:
\[ 10u = 450 \]
\[ u = 45 \, \text{m/s} \]

### Step 2: Calculate the time taken to reach point A (20 m below the tower)
For point A, the distance (s) is 20 m below the tower:
\[ s = -30 \, \text{m} \] (since it is 50 m - 20 m)

Using the same equation of motion:
\[ -30 = 45t_1 - \frac{1}{2}(10)t_1^2 \]
\[ -30 = 45t_1 - 5t_1^2 \]
Rearranging gives:
\[ 5t_1^2 - 45t_1 - 30 = 0 \]
Dividing through by 5:
\[ t_1^2 - 9t_1 - 6 = 0 \]

Using the quadratic formula:
\[ t_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -9, c = -6 \):
\[ t_1 = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \]
\[ t_1 = \frac{9 \pm \sqrt{81 + 24}}{2} \]
\[ t_1 = \frac{9 \pm \sqrt{105}}{2} \]
Calculating gives:
\[ t_1 \approx 9.4 \, \text{s} \]

### Step 3: Calculate the time taken to reach point B (40 m below the tower)
For point B, the distance (s) is 40 m below the tower:
\[ s = -10 \, \text{m} \] (since it is 50 m - 40 m)

Using the same equation of motion:
\[ -10 = 45t_2 - \frac{1}{2}(10)t_2^2 \]
\[ -10 = 45t_2 - 5t_2^2 \]
Rearranging gives:
\[ 5t_2^2 - 45t_2 - 10 = 0 \]
Dividing through by 5:
\[ t_2^2 - 9t_2 - 2 = 0 \]

Using the quadratic formula:
\[ t_2 = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]
\[ t_2 = \frac{9 \pm \sqrt{81 + 8}}{2} \]
\[ t_2 = \frac{9 \pm \sqrt{89}}{2} \]
Calculating gives:
\[ t_2 \approx 9.8 \, \text{s} \]

### Step 4: Calculate the time taken to cover distance AB
The time taken to cover distance AB is:
\[ t_{AB} = t_2 - t_1 \]
\[ t_{AB} = 9.8 - 9.4 = 0.4 \, \text{s} \]

### Final Answer
The time taken to cover the distance AB is **0.4 seconds**.