Question

Two bodies are thrown simultaneously from a tower with same initial velocity $v_0$ one vertically upwards, the other vertically downwards. The distance between the two bodies after time t is.

• A. $2v_{0}t+\frac{1}{2}g{t}^2$
• B. $2v_{0}t$
• C. $v_{0}t+\frac{1}{2}g{t}^2$
• D. $v_{0}t$

Answer 1

The correct option is B $2v_{0}t$

For vertically upward motion,

$h_1=v_{0}t-\frac{1}{2}g{t}^2$

and for vertically downward motion,

$h_0=v_{0}t+\frac{1}{2}g{t}^2$

$\therefore$ Total distance covered in $t$ sec

$h=h_1+h_2=2v_{0}t$