Two balls of equal mass are projected upwards simultaneously, one from the ground with initial velocity `50ms^-1` and the other from a `40m` tower with initial velocity of `30ms^-1`. The maximum height attained by their COM will be a) 80 m b) 60 m c) 100 m d) 120 m

Options

A. (a) `80m`
B. (b) `60m`
C. (c) `100m`
D. (d) `120m`

Explanation

To solve the problem, we need to calculate the maximum height attained by the center of mass (COM) of the two balls. Let's break down the solution step by step.

### Step 1: Identify the Initial Positions of the Balls
- Ball 1 is projected from the ground with an initial velocity of \(50 \, \text{m/s}\).
- Ball 2 is projected from a height of \(40 \, \text{m}\) with an initial velocity of \(30 \, \text{m/s}\).

### Step 2: Calculate the Initial Position of the Center of Mass (COM)
The formula for the position of the center of mass \(X_{CM}\) of two objects is given by:
\[ X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
Since both balls have equal mass \(m\):
- For Ball 1 (at ground level): \(x_1 = 0 \, \text{m}\)
- For Ball 2 (at height of 40 m): \(x_2 = 40 \, \text{m}\)

Thus, the initial position of the center of mass is:
\[ X_{CM} = \frac{m \cdot 0 + m \cdot 40}{m + m} = \frac{40m}{2m} = 20 \, \text{m} \]

### Step 3: Calculate the Initial Velocity of the Center of Mass (COM)
The initial velocity of the center of mass \(V_{CM}\) is given by:
\[ V_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \]
Where:
- \(v_1 = 50 \, \text{m/s}\) (velocity of Ball 1)
- \(v_2 = 30 \, \text{m/s}\) (velocity of Ball 2)

Thus, we have:
\[ V_{CM} = \frac{m \cdot 50 + m \cdot 30}{m + m} = \frac{80m}{2m} = 40 \, \text{m/s} \]

### Step 4: Calculate the Maximum Height Reached by the Center of Mass
Using the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \(v = 0 \, \text{m/s}\) (final velocity at maximum height)
- \(u = V_{CM} = 40 \, \text{m/s}\)
- \(a = -g = -10 \, \text{m/s}^2\) (acceleration due to gravity)

Setting up the equation:
\[ 0 = (40)^2 + 2(-10)s \]
\[ 0 = 1600 - 20s \]
\[ 20s = 1600 \]
\[ s = \frac{1600}{20} = 80 \, \text{m} \]

### Step 5: Calculate the Total Maximum Height of the Center of Mass from the Ground
The total maximum height of the center of mass from the ground level is:
\[ \text{Total Height} = \text{Initial Height of COM} + \text{Height Gained} \]
\[ \text{Total Height} = 20 \, \text{m} + 80 \, \text{m} = 100 \, \text{m} \]

### Final Answer
The maximum height attained by the center of mass is \(100 \, \text{m}\). Therefore, the correct option is:
**c) 100 m**

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