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 X and Y. Let's denote the initial speed of both balls as \( v_0 \), the height of the tower as \( h \), and the acceleration due to gravity as \( g = 10 \, \text{m/s}^2 \).

### Step 1: Analyze the motion of ball X (thrown upward)
For ball X, which is thrown upwards, the time taken to reach the ground is \( t_X = 6 \, \text{s} \). The equation of motion can be expressed as:

\[ h = v_0 t_X - \frac{1}{2} g t_X^2 \]

Substituting the values we have:

\[ h = v_0 (6) - \frac{1}{2} (10) (6^2) \]

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

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

Thus, the equation becomes:

\[ h = 6v_0 - 180 \quad \text{(1)} \]

### Step 2: Analyze the motion of ball Y (thrown downward)
For ball Y, which is thrown downwards, the time taken to reach the ground is \( t_Y = 2 \, \text{s} \). The equation of motion can be expressed as:

\[ h = v_0 t_Y + \frac{1}{2} g t_Y^2 \]

Substituting the values we have:

\[ h = v_0 (2) + \frac{1}{2} (10) (2^2) \]

Calculating \( \frac{1}{2} (10) (4) \):

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

Thus, the equation becomes:

\[ h = 2v_0 + 20 \quad \text{(2)} \]

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

\[ 6v_0 - 180 = 2v_0 + 20 \]

### Step 4: Solve for \( v_0 \)
Rearranging the equation gives:

\[ 6v_0 - 2v_0 = 180 + 20 \]

\[ 4v_0 = 200 \]

\[ v_0 = \frac{200}{4} = 50 \, \text{m/s} \]

### Step 5: Substitute \( v_0 \) back to find \( h \)
Now, we can substitute \( v_0 \) back into either equation (1) or (2) to find the height \( h \). Using equation (1):

\[ h = 6(50) - 180 \]

\[ h = 300 - 180 = 120 \, \text{m} \]

### Final Answers
The initial speed of each ball is \( v_0 = 50 \, \text{m/s} \) and the height of the tower is \( h = 120 \, \text{m} \).

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