There is a tower of height 20m and a particle is projected up from top of the tower with an initial speed 20m//s. Top of the tower is marked as point A, from where particle is projected. Point of maximum height a denoted as B. When particle reaches the point A during downward journey then we call the same point as C. Point at the bottom of tower is marked as D where particle finally strikes. Acceleration due to gravity `g=10m//s^(2)`
Time taken by the particle to reach the ground is

Options

A. 2s
B. 4s
C. `2sqrt(2s)`
D. `(2+2sqrt(2))s`

Explanation

To solve the problem of a particle projected from the top of a tower, we will follow these steps:

### Step 1: Understand the problem
We have a tower of height \( h = 20 \, \text{m} \) and a particle is projected upwards from the top of the tower with an initial speed \( u = 20 \, \text{m/s} \). The acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \). We need to find the time taken for the particle to reach the ground (point D).

### Step 2: Set up the equation of motion
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s \) is the displacement,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( t \) is the time.

### Step 3: Define the signs
Since we are considering the downward direction as negative:
- Displacement \( s = -20 \, \text{m} \) (from point A to point D),
- Initial velocity \( u = 20 \, \text{m/s} \) (upwards, so positive),
- Acceleration \( a = -10 \, \text{m/s}^2 \) (downwards, due to gravity).

### Step 4: Substitute values into the equation
Substituting the values into the equation:
\[ -20 = 20t - \frac{1}{2}(10)t^2 \]
This simplifies to:
\[ -20 = 20t - 5t^2 \]
Rearranging gives us:
\[ 5t^2 - 20t - 20 = 0 \]

### Step 5: Simplify the equation
Dividing the entire equation by 5:
\[ t^2 - 4t - 4 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -4, c = -4 \):
\[ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ t = \frac{4 \pm \sqrt{16 + 16}}{2} \]
\[ t = \frac{4 \pm \sqrt{32}}{2} \]
\[ t = \frac{4 \pm 4\sqrt{2}}{2} \]
\[ t = 2 \pm 2\sqrt{2} \]

### Step 7: Determine the positive time
Since time cannot be negative, we take the positive solution:
\[ t = 2 + 2\sqrt{2} \, \text{seconds} \]

### Final Answer
The time taken by the particle to reach the ground is \( t = 2 + 2\sqrt{2} \, \text{seconds} \).

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