Three particles A, B and C are thrown from the top of a tower with the same speed. A is thrown up, B is thrown down and C is horizontally. They hit the ground with speeds `v_(A),v_(B) and v_(C)` respectively then,

Options

A. `V_(A) gt V_(B) gtV_(C )`
B. `V_(A) lt V_(B) lt V_(C )`
C. `V_(A) = V_(B) gt V_(C )`
D. `V_(A) =V_(B) = V_(C )`

Answer

D

Explanation

In all three cases we can apply conservation of mechanical energy to calculate speed at the ground level .
`1/2 mv^(2) - 1/2"mu"^(2) = mgh rArr v = sqrt((u^(2)+2gh))`
We get the same speed at the ground level .