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 (1/2)m from the ground, it explodes into a number of pieces which follow different parabolic parts. What is the velocity of centre of mass, when it is at a height of 1 m from the ground? `(g=10ms^(-2))`

Options

A. `4sqrt(5)ms^(-1)`
B. `2sqrt(5)ms^(-1)`
C. `5sqrt(4)ms^(-1)`
D. `10ms^(-1)`

Answer

A

Explanation

Motion of centre of mass is exactly similar to that of translatory motionn of a body that is thrown into air.
`u_(x)=ucostheta" "u_(y)=usintheta" "=10/sqrt(2)m//s`
`v_(x)=10/sqrt(2)m//s` (since there is no change in the horizontal velocity)
`v_(y)^(2)-u_(y)^(2)=2(-g)(h), " "v_(y)^(2)=u_(y)^(2)-2gh=100/2-2 times 10 times 1=30`
`therefore` Net velocity of CM `=sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(100/2+30)=sqrt(80)=4sqrt(5)m//s`