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 paths. What is the velocity of centre of mass, when it is at a height of 1 m from the ground? `(g=10 ms^(-2))`

Options

A. `4sqrt5ms^(-1)`
B. `2sqrt5ms^(-1)`
C. `5sqrt4ms^(-1)`
D. `10 ms^(-1)`

Answer

A

Explanation

Motion of centre of mass is exactly similarly that translatory motion of a body that is thrown into air.
`u_x=ucostheta" "u_y=usintheta=10/sqrt2m//s`
`v_x=10/sqrt2m//s` (since there is no change iri the horizontal velocity)
`v_y^2-u_y^2=2(-g)(h),v_y^2=u_y^2-2gh=100/2-2xx10xx1=30`
`:.` Net velocity of CM `CM=sqrt(v_x^2+v_y^2)=sqrt(100/2+30)=sqrt80=4sqrt5m//s `