Two-body meeting problem – one dropped from 60 m, the other projected upward A particle is dropped from the top of a 60 m tower. At the same instant, another particle is projected vertically upward from the base so that both meet at a height of 15.9 m above the ground. Find the initial upward speed of the second particle (use g = 9.8 m/s^2).

Introduction / Context:
Relative-motion problems with constant acceleration are common in kinematics. Here, two particles start simultaneously from opposite ends and meet at a specified point. The solution uses basic constant-acceleration equations without calculus.

Given Data / Assumptions:

• Height of tower H = 60 m.
• Meeting point height from ground y = 15.9 m.
• Gravity g = 9.8 m/s^2; air resistance neglected.
• First particle drops from rest at the top; second is launched upward from ground with speed u.

Concept / Approach:

Let both start at t = 0 and meet after time t. Use vertical motion equations for each particle and set their positions equal to the meeting height to solve for t and then u.

Step-by-Step Solution:

Verification / Alternative check:

Using relative motion: the relative speed at the meeting instant equals the sum of their instantaneous speeds; time computed above is consistent with both trajectories reaching 15.9 m simultaneously.

Why Other Options Are Wrong:

16, 18, and 22 m/s do not satisfy the kinematic equation at y = 15.9 m when t = 3 s. 24 m/s overshoots the meeting height too early.

Common Pitfalls:

Using g = 10 m/s^2 without adjusting heights; mixing sign conventions; miscomputing distance fallen versus height above ground.

Final Answer:

20 m/s