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

Options

A. `2v_(0)t+(1)/(2) "gt"^(2)`
B. `2v_(0) t `
C. `v_(0) t+(1)/(2) "gt"^(2)`
D. `v_(0) t `

Answer

B

Explanation

For veetically upward motion, `h_(1)=v_(0)t-(1)/(2)"gt"^(2)`
and for vertically downward motion,
`h_(2)=v_(0)t+(1)/(2) g t^(2)`
`therefore` Total distance covered in t sec
`h=h_(1)+h_(2)=2v_(0) t.`