A cracker is thrown into air with a velocity of `10 m//s` at an angle of `45^(@)` with the vertical. When it is at a height of `0.5 m` from the ground, it explodes into a number of pieces which follow different parabolic paths. What is the velocity of centre of mass, when it is at a height of `1 m` from the ground? (`g = 10 m//s^(2)`)

Options

A. `4sqrt(5)ms^(-1)`
B. `2sqrt(5)ms^(-1)`
C. `5sqrt(4)ms^(-1)`
D. `5ms^(-1)`

Answer

A

Explanation

During explosion of a projectile, path of center of mass does not change. At height 1m,
`v = sqrt(u^(2)-2gh)` (From conservation of mechanical energy)
`= 4sqrt(5)ms^(-1)`