From the top of a tower, a stone is thrown up and reaches the ground in time `t_(1)=9`s. a second stone is thrown down with the same speed and reaches the ground in time `t_(2)=4s`. A third stone is released from rest and reaches the ground in time `t_(3)`, which is equal to

Options

A. 6.5 s
B. 6.0 s
C. `(72)/(13)s`
D. none

Answer

To solve the problem, we need to analyze the motion of three stones thrown from the top of a tower. Let's break down the solution step by step.

### Step 1: Analyze the first stone thrown upwards
- The first stone is thrown upwards with an initial velocity \( u \) and takes time \( t_1 = 9 \) seconds to reach the ground.
- The stone first goes up, then comes down to the level of the tower, and finally falls to the ground.
- The total time of flight can be split into two parts: the time to reach the maximum height \( t_{up} \) and the time to fall back down to the ground \( t_{down} \).

### Step 2: Establish equations for the first stone
- The time to go up and come back down to the level of the tower can be expressed as:
\[ t_{up} + t_{down} = t_1 \]
- The time taken to fall from the top of the tower to the ground can be expressed using the equation of motion:
\[ h = u t_{down} + \frac{1}{2} g t_{down}^2 \]

### Step 3: Analyze the second stone thrown downwards
- The second stone is thrown downwards with the same initial velocity \( u \) and takes time \( t_2 = 4 \) seconds to reach the ground.
- The equation of motion for the second stone is:
\[ h = u t_2 + \frac{1}{2} g t_2^2 \]

### Step 4: Relate the two cases
- From the first case, we can find the time taken to fall from the level of the tower to the ground:
\[ t_{down} = t_1 - t_{up} \]
- From the second case, we can express the height \( h \) in terms of \( u \) and \( g \):
\[ h = u t_2 + \frac{1}{2} g t_2^2 \]

### Step 5: Calculate the height of the tower
- Using the second stone's equation:
\[ h = u \cdot 4 + \frac{1}{2} \cdot 10 \cdot 4^2 \]
- Substitute \( g = 10 \, \text{m/s}^2 \):
\[ h = 4u + 80 \]

### Step 6: Calculate the initial velocity \( u \)
- From the first stone's motion, we can also express \( h \):
\[ h = u (t_1 - t_{up}) + \frac{1}{2} g (t_1 - t_{up})^2 \]
- We can find \( u \) by equating the two expressions for \( h \).

### Step 7: Analyze the third stone released from rest
- The third stone is released from rest and falls freely under gravity. The time taken to reach the ground is \( t_3 \).
- The equation of motion for the third stone is:
\[ h = \frac{1}{2} g t_3^2 \]

### Step 8: Solve for \( t_3 \)
- We already have \( h \) from previous calculations. Substitute \( h \) into the equation:
\[ 180 = \frac{1}{2} \cdot 10 \cdot t_3^2 \]
- Simplifying gives:
\[ 180 = 5 t_3^2 \implies t_3^2 = \frac{180}{5} = 36 \implies t_3 = 6 \, \text{s} \]

### Final Answer
Thus, the time \( t_3 \) for the third stone is:
\[ \boxed{6 \, \text{s}} \]