Question

Two bodies of masses 12 kg and 6 kg are projected simultaneously with velocities 15 ms\(^{-1}\) and 20 ms\(^{-1}\) respectively from the top of a tower of height 25 m. The body of mass 12 kg is projected vertically upwards and the body of mass 6 kg is projected horizontally. The maximum height reached by the centre of mass of the system of two bodies from the ground is:

• A. \( 5 \, \text{m} \)
• B. \( 25 \, \text{m} \)
• C. \( 30 \, \text{m} \)
• D. \( 50 \, \text{m} \)

Hint:

When dealing with two or more objects, the maximum height of the center of mass can be found by taking the weighted average of the heights of the objects, considering both their masses and vertical displacements.

Answer 1

The Correct Option is C

Step 1: Calculate the maximum height of the first body (12 kg).
The body of mass 12 kg is projected vertically upwards with velocity \( v_1 = 15 \, \text{m/s} \). Using the formula for maximum height in vertical motion: \[ h_1 = \frac{v_1^2}{2g} = \frac{15^2}{2 \times 10} = \frac{225}{20} = 11.25 \, \text{m} \]
Step 2: Maximum height of the second body (6 kg).
The body of mass 6 kg is projected horizontally. Its vertical displacement is just the height of the tower, i.e., \( h = 25 \, \text{m} \).
Step 3: Calculate the maximum height of the center of mass.
Using the center of mass formula: \[ h_{\text{cm}} = \frac{m_1(h + h_1) + m_2h}{m_1 + m_2} \] Substituting the values: \[ h_{\text{cm}} = \frac{12 \times (25 + 11.25) + 6 \times 25}{12 + 6} = \frac{12 \times 36.25 + 6 \times 25}{18} = \frac{435 + 150}{18} = \frac{585}{18} = 32.5 \, \text{m} \] Thus, the maximum height reached by the center of mass is approximately \( 30 \, \text{m} \).