Question

Two bodies A and B are projected simultaneously with velocities \( 20 \, \text{ms}^{-1} \) and \( 40 \, \text{ms}^{-1} \), respectively. Body A is projected vertically up from the top of a tower of height 80 m and body B is projected vertically up from the bottom of the same tower. The bodies A and B meet in time of:

• A. 5 s
• B. 3 s
• C. 6 s
• D. 4 s

Hint:

For two bodies moving towards each other, solve the equations of motion for each body and find when their displacements add up to the total distance.

Answer 1

The Correct Option is D

- For body A, which is projected upwards, use the equation of motion: \[ h_A = v_0 t + \frac{1}{2} (-g) t^2 \] where \( h_A = 80 \, \text{m} \), \( v_0 = 20 \, \text{ms}^{-1} \), and \( g = 10 \, \text{ms}^{-2} \).
- For body B, the motion is also vertical but downwards, and it will follow: \[ h_B = v_0 t + \frac{1}{2} g t^2 \] where \( v_0 = 40 \, \text{ms}^{-1} \).
- Solving these two equations for the time \( t \), the bodies will meet after 4 seconds.