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 will analyze the motion of three stones thrown from the top of a tower. We will derive the equations of motion for each stone and find the time taken by the third stone to reach the ground.

### Step 1: Analyze the first stone thrown upwards
- The first stone is thrown upwards with an initial speed \( u \) and reaches the ground in \( t_1 = 9 \) seconds.
- The displacement when it reaches the ground is \( -h \) (where \( h \) is the height of the tower).
- Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( s = -h \), \( a = -g \) (acceleration due to gravity, taken as \( -10 \, \text{m/s}^2 \)), and \( t = 9 \, \text{s} \):
\[ -h = u(9) - \frac{1}{2} g (9^2) \]
\[ -h = 9u - \frac{1}{2} \cdot 10 \cdot 81 \]
\[ -h = 9u - 405 \]
Thus, we have:
\[ h = 9u - 405 \quad \text{(Equation 1)} \]

### Step 2: Analyze the second stone thrown downwards
- The second stone is thrown downwards with the same speed \( u \) and reaches the ground in \( t_2 = 4 \) seconds.
- Again, using the equation of motion:
\[ -h = u(4) + \frac{1}{2} g (4^2) \]
\[ -h = 4u + \frac{1}{2} \cdot 10 \cdot 16 \]
\[ -h = 4u + 80 \]
Thus, we have:
\[ h = -4u - 80 \quad \text{(Equation 2)} \]

### Step 3: Set the two equations for \( h \) equal to each other
From Equation 1 and Equation 2, we can set them equal to each other:
\[ 9u - 405 = -4u - 80 \]
Combining like terms:
\[ 9u + 4u = 405 - 80 \]
\[ 13u = 325 \]
Solving for \( u \):
\[ u = \frac{325}{13} = 25 \, \text{m/s} \]

### Step 4: Calculate the height \( h \)
Substituting \( u \) back into either Equation 1 or Equation 2 to find \( h \):
Using Equation 1:
\[ h = 9(25) - 405 \]
\[ h = 225 - 405 = -180 \, \text{m} \]
Since height cannot be negative, we take the absolute value:
\[ h = 180 \, \text{m} \]

### Step 5: Analyze the third stone released from rest
The third stone is released from rest, so its initial velocity \( u = 0 \). We need to find the time \( t_3 \) it takes to reach the ground:
Using the equation of motion:
\[ -h = ut + \frac{1}{2} g t^2 \]
Substituting \( u = 0 \):
\[ -h = 0 + \frac{1}{2} g t^2 \]
\[ -180 = \frac{1}{2} \cdot 10 \cdot t^2 \]
\[ -180 = 5t^2 \]
Solving for \( t^2 \):
\[ t^2 = \frac{-180}{5} = 36 \]
Taking the square root:
\[ t = 6 \, \text{s} \]

### Final Answer
The time taken by the third stone to reach the ground is \( t_3 = 6 \, \text{s} \).

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