Two balls X and Y are thrown from top of tower one vertically upward and other vertically downward with same speed. If times taken by them to reach the ground are 6 s and 2 s respectively, then the height of the tower and initial speed of each ball are `(g=10m//s^(2))`

Options

A. 60 m, 15 m/s
B. 80 m, 20 m/s
C. 60 m, 20 m/s
D. 45 m, 10 m/s

Answer

To solve the problem, we need to analyze the motion of both balls thrown from the top of the tower. Let's denote the height of the tower as \( h \) and the initial speed of each ball as \( u \). The acceleration due to gravity is given as \( g = 10 \, \text{m/s}^2 \).

### Step 1: Analyze the motion of the ball thrown upward (Ball X)
For the ball thrown vertically upward, the time taken to reach the ground is \( t_1 = 6 \, \text{s} \). The equation of motion for this ball is given by:

\[ h = -ut_1 + \frac{1}{2}gt_1^2 \]

Substituting \( t_1 = 6 \, \text{s} \) and \( g = 10 \, \text{m/s}^2 \):

\[ h = -u(6) + \frac{1}{2}(10)(6^2) \]

Calculating \( \frac{1}{2}(10)(6^2) \):

\[ \frac{1}{2}(10)(36) = 180 \]

Now substituting this back into the equation:

\[ h = -6u + 180 \quad \text{(Equation 1)} \]

### Step 2: Analyze the motion of the ball thrown downward (Ball Y)
For the ball thrown vertically downward, the time taken to reach the ground is \( t_2 = 2 \, \text{s} \). The equation of motion for this ball is:

\[ h = ut_2 + \frac{1}{2}gt_2^2 \]

Substituting \( t_2 = 2 \, \text{s} \) and \( g = 10 \, \text{m/s}^2 \):

\[ h = u(2) + \frac{1}{2}(10)(2^2) \]

Calculating \( \frac{1}{2}(10)(2^2) \):

\[ \frac{1}{2}(10)(4) = 20 \]

Now substituting this back into the equation:

\[ h = 2u + 20 \quad \text{(Equation 2)} \]

### Step 3: Set the two equations equal to each other
Since both equations represent the same height \( h \), we can set them equal to each other:

\[ -6u + 180 = 2u + 20 \]

### Step 4: Solve for \( u \)
Rearranging the equation to isolate \( u \):

\[ 180 - 20 = 2u + 6u \]

\[ 160 = 8u \]

Dividing both sides by 8:

\[ u = \frac{160}{8} = 20 \, \text{m/s} \]

### Step 5: Substitute \( u \) back to find \( h \)
Now that we have \( u \), we can substitute it back into either Equation 1 or Equation 2 to find \( h \). We will use Equation 2:

\[ h = 2u + 20 \]

Substituting \( u = 20 \):

\[ h = 2(20) + 20 = 40 + 20 = 60 \, \text{m} \]

### Final Answers
- The height of the tower \( h = 60 \, \text{m} \)
- The initial speed of each ball \( u = 20 \, \text{m/s} \)