Question

Two particles A and B are thrown vertically upward with velocity, 5m/s and 10 m/s respectively ($g=10m/s^2$). Find separation between them after one second.

• A. 1 m
• B. 2 m
• C. 5 m
• D. 10 m

Answer 1

The correct option is C

5 m

Method 1:
Position of A after 1 s, $s_A=ut-\frac{1}{2}g{t}^2=5\times 1-\frac{1}{2}\times 10\times 1^2=0$
i.e., the particle will return to ground at t = 1 s.
The position of B after 1 s, $s_B=ut-\frac{1}{2}g{t}^2=10\times 1-\frac{1}{2}\times 10\times 1^2=5m$
Hence, separation between A and B, $s_B-s_A=5m$
Method 2: Relative velocity method
Acceleration of B w.r.t A $=g-g=0$
Initial relative velocity $=10-5=5m/s$
Hence, relative separation between particles $5\times 1=5m$