From the top of a tower of height 40m, a ball is projected upward with a speed of `20ms^(-1)` at an angle of elevation of `30^(@)`. Then the ratio of the total time taken by the ball to hit the ground to the time taken to ball come at same level as top of tower.

Options

A. `2:1`
B. `3:1`
C. `3:2`
D. `4:1`

Explanation

To solve the problem step by step, we need to find the total time taken by the ball to hit the ground (T1) and the time taken by the ball to reach the same level as the top of the tower (T2). We will then find the ratio T1/T2.

### Step 1: Break down the initial velocity into components
The ball is projected with a speed of 20 m/s at an angle of 30 degrees. We can find the vertical and horizontal components of the initial velocity (U).

- **Vertical component (U_y)**:
\[ U_y = U \sin(\theta) = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]

- **Horizontal component (U_x)**:
\[ U_x = U \cos(\theta) = 20 \cos(30^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

### Step 2: Calculate time taken to reach the same level as the top of the tower (T2)
To find T2, we need to determine how long it takes for the ball to come back to the height of the tower (40 m).

Using the equation of motion in the vertical direction:
\[ S = U_y t - \frac{1}{2} g t^2 \]
Where:
- \(S = -40 \, \text{m}\) (the ball falls 40 m)
- \(U_y = 10 \, \text{m/s}\)
- \(g = 10 \, \text{m/s}^2\)

Substituting the values:
\[ -40 = 10t - \frac{1}{2} \cdot 10 t^2 \]
\[ -40 = 10t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 10t - 40 = 0 \]
Dividing through by 5:
\[ t^2 - 2t - 8 = 0 \]

### Step 3: Solve the quadratic equation for T2
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 1\), \(b = -2\), and \(c = -8\):
\[ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \]
\[ t = \frac{2 \pm \sqrt{4 + 32}}{2} \]
\[ t = \frac{2 \pm \sqrt{36}}{2} \]
\[ t = \frac{2 \pm 6}{2} \]
This gives us two possible solutions:
\[ t = \frac{8}{2} = 4 \, \text{s} \quad \text{and} \quad t = \frac{-4}{2} = -2 \, \text{s} \, (\text{not valid}) \]
Thus, \(T2 = 4 \, \text{s}\).

### Step 4: Calculate total time taken to hit the ground (T1)
The total time (T1) is the time taken to reach the maximum height plus the time taken to fall from the maximum height to the ground.

1. **Time to reach maximum height (T_up)**:
Using \(V = U_y - g t\) where \(V = 0\) at maximum height:
\[ 0 = 10 - 10 T_{up} \implies T_{up} = 1 \, \text{s} \]

2. **Total time to fall from maximum height to ground (T_down)**:
The maximum height can be calculated as:
\[ H = U_y T_{up} - \frac{1}{2} g T_{up}^2 = 10 \cdot 1 - \frac{1}{2} \cdot 10 \cdot 1^2 = 10 - 5 = 5 \, \text{m} \]
The total height from which the ball falls is \(40 + 5 = 45 \, \text{m}\).

Using the equation of motion again for the downward journey:
\[ S = \frac{1}{2} g T_{down}^2 \]
\[ 45 = \frac{1}{2} \cdot 10 T_{down}^2 \implies 45 = 5 T_{down}^2 \implies T_{down}^2 = 9 \implies T_{down} = 3 \, \text{s} \]

Thus, the total time \(T1\) is:
\[ T1 = T_{up} + T_{down} = 1 + 3 = 4 \, \text{s} \]

### Step 5: Calculate the ratio T1/T2
Now we can find the ratio:
\[ \text{Ratio} = \frac{T1}{T2} = \frac{4 \, \text{s}}{2 \, \text{s}} = 2 \]

### Final Answer
The ratio of the total time taken by the ball to hit the ground to the time taken to reach the same level as the top of the tower is **2**.