A stone of mass 500g is dropped from the top of a tower of 100m height. Simultaneously another stone of mass 1 kg is thrown horizontally with a speed of `10 m s^(-1)` from same point. The height of the centre of mass of the above two stone system after 3 sec is `( g = 10 m s^(-2)`)

Options

A. 45 m
B. 35 m
C. 55 m
D. none of these

Answer

To solve the problem, we need to find the height of the center of mass of the two stones after 3 seconds. We will analyze the motion of both stones separately and then calculate the center of mass.

### Step 1: Analyze the motion of the first stone (mass = 500 g)
The first stone is dropped from a height of 100 m. The initial velocity (u) is 0 m/s, and the acceleration (a) is due to gravity, which is -10 m/s² (negative because it is downward).

Using the equation of motion:
\[ S_y = u_y t + \frac{1}{2} a_y t^2 \]

Substituting the values:
\[ S_y = 0 \cdot 3 + \frac{1}{2} \cdot (-10) \cdot (3^2) \]
\[ S_y = 0 - 5 \cdot 9 \]
\[ S_y = -45 \text{ m} \]

This means that after 3 seconds, the first stone has fallen 45 m from the top of the tower.

### Step 2: Calculate the height of the first stone from the ground
The initial height of the tower is 100 m. Thus, the height of the first stone from the ground after 3 seconds is:
\[ \text{Height from ground} = 100 - 45 = 55 \text{ m} \]

### Step 3: Analyze the motion of the second stone (mass = 1 kg)
The second stone is thrown horizontally with a speed of 10 m/s. Initially, it has no vertical velocity (u = 0 m/s) and is also subjected to the same gravitational acceleration of -10 m/s².

Using the same equation of motion for vertical displacement:
\[ S_y = u_y t + \frac{1}{2} a_y t^2 \]
\[ S_y = 0 \cdot 3 + \frac{1}{2} \cdot (-10) \cdot (3^2) \]
\[ S_y = 0 - 5 \cdot 9 \]
\[ S_y = -45 \text{ m} \]

Thus, after 3 seconds, the second stone has also fallen 45 m.

### Step 4: Calculate the height of the second stone from the ground
The height of the second stone from the ground after 3 seconds is also:
\[ \text{Height from ground} = 100 - 45 = 55 \text{ m} \]

### Step 5: Calculate the center of mass of the two stones
The center of mass (CM) of the two stones can be calculated using the formula:
\[ h_{CM} = \frac{m_1 h_1 + m_2 h_2}{m_1 + m_2} \]
Where:
- \( m_1 = 0.5 \text{ kg} \) (mass of the first stone)
- \( h_1 = 55 \text{ m} \) (height of the first stone)
- \( m_2 = 1 \text{ kg} \) (mass of the second stone)
- \( h_2 = 55 \text{ m} \) (height of the second stone)

Substituting the values:
\[ h_{CM} = \frac{(0.5 \times 55) + (1 \times 55)}{0.5 + 1} \]
\[ h_{CM} = \frac{27.5 + 55}{1.5} \]
\[ h_{CM} = \frac{82.5}{1.5} \]
\[ h_{CM} = 55 \text{ m} \]

### Final Answer
The height of the center of mass of the two stone system after 3 seconds is **55 m**.

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