A stone is thrown vertically upwards with an initial velocity u form the top of a tower, recahes the ground with a velocity 3u. The height of the tower is

Options

A. `(3u^(2))/(g)`
B. `(4u^(2))/(g)`
C. `(6u^(2))/(g)`
D. `(9u^(2))/(g)`

Explanation

To solve the problem of finding the height of the tower from which a stone is thrown vertically upwards with an initial velocity \( u \) and reaches the ground with a final velocity of \( 3u \), we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial velocity (\( u \)) = \( u \)
- Final velocity (\( v \)) = \( 3u \)
- Acceleration (\( a \)) = \( -g \) (since the stone is moving against gravity when thrown upwards)
- Displacement (\( s \)) = height of the tower (\( h \))

2. **Use the Kinematic Equation:**
The kinematic equation we will use is:
\[ v^2 = u^2 + 2as \]
Here, we will substitute the values we have:
\[ (3u)^2 = u^2 + 2(-g)(h) \]

3. **Substitute and Simplify:**
- Substitute \( v = 3u \) and \( u = u \):
\[ 9u^2 = u^2 - 2gh \]
- Rearranging gives:
\[ 9u^2 - u^2 = -2gh \]
\[ 8u^2 = -2gh \]

4. **Solve for Height (\( h \)):**
- Rearranging the equation to solve for \( h \):
\[ h = \frac{8u^2}{2g} \]
- Simplifying further:
\[ h = \frac{4u^2}{g} \]

5. **Final Answer:**
The height of the tower is:
\[ h = \frac{4u^2}{g} \]