A stone is thrown vertically upwards with an initial speed `u` from the top of a tower, reaches the ground with a speed `3 u`. 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 find the height of the tower from which a stone is thrown vertically upwards with an initial speed \( u \) and reaches the ground with a speed of \( 3u \), we can use the kinematic equations of motion. Here’s a step-by-step solution:

### Step 1: Understand the motion of the stone
When the stone is thrown upwards, it first moves up until it reaches its maximum height, then it falls back down to the ground. The initial velocity when thrown is \( u \), and the final velocity when it hits the ground is \( 3u \).

### Step 2: Apply the kinematic equation
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity = \( 3u \)
- \( u \) = initial velocity = \( u \)
- \( a \) = acceleration due to gravity = \( -g \) (since it's acting downwards)
- \( s \) = height of the tower = \( h \)

### Step 3: Substitute the values into the equation
Since we are considering downward direction as positive, we can write the equation as:
\[ (3u)^2 = u^2 + 2(-g)(h) \]
This simplifies to:
\[ 9u^2 = u^2 - 2gh \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ 9u^2 - u^2 = -2gh \]
\[ 8u^2 = -2gh \]

### Step 5: Solve for height \( h \)
Now, we can solve for \( h \):
\[ h = \frac{8u^2}{2g} = \frac{4u^2}{g} \]

### Conclusion
Thus, the height of the tower is:
\[ h = \frac{4u^2}{g} \]