A stone is thrown vertically upwards with an initial velocity u from the top of a tower of lieight `(12u^(2))/(g)` With what velocity does the stone reach the ground?

Options

A. u
B. 4u
C. 5u
D. `2sqrt(6)u`

Explanation

To solve the problem, we need to determine the final velocity of a stone thrown vertically upwards from the top of a tower when it reaches the ground. The height of the tower is given as \( h = \frac{12u^2}{g} \), where \( u \) is the initial velocity of the stone and \( g \) is the acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- The initial velocity of the stone, \( u \).
- The height of the tower, \( h = \frac{12u^2}{g} \).
- The acceleration due to gravity, \( g \).

2. **Determine the maximum height reached by the stone**:
When the stone is thrown upwards, it will first ascend until it reaches its maximum height. The maximum height \( H \) can be calculated using the formula:
\[ H = \frac{u^2}{2g} \]
However, since the stone is thrown from a height \( h \), the total height from the ground to the maximum point is:
\[ H_{total} = h + H = \frac{12u^2}{g} + \frac{u^2}{2g} = \frac{12u^2 + 0.5u^2}{g} = \frac{12.5u^2}{g} \]

3. **Use the kinematic equation to find the final velocity**:
We can use the third equation of motion, which relates initial velocity, final velocity, acceleration, and displacement:
\[ v^2 = u^2 + 2gh \]
Here, \( v \) is the final velocity when the stone reaches the ground, \( u \) is the initial velocity (upwards), and \( h \) is the total height from which the stone falls back to the ground.

Substituting the values:
\[ v^2 = u^2 + 2g\left(\frac{12u^2}{g}\right) \]
Simplifying this gives:
\[ v^2 = u^2 + 24u^2 = 25u^2 \]

4. **Calculate the final velocity**:
Taking the square root of both sides:
\[ v = \sqrt{25u^2} = 5u \]

### Final Answer:
The velocity with which the stone reaches the ground is \( v = 5u \).

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