A cracker is thrown into air with a velocity of `10ms^(-1)` 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 center of mass, when it is at a height of 1m from the ground? (Take, `g = 10 ms^(-2)`]

Options

A. `4sqrt(5)m//s`
B. `2sqrt(5)m//s`
C. `5sqrt(4)m//s`
D. `10m//s`

Answer

A

Explanation

Motion of centre of mass is exactly similar to the motion of a body had it not exploded.
`u_(x)=ucostheta=10/(sqrt(2))m//s, u_(y)=usintheta=10/(sqrt(2))m//s`
`v_(x)=u_(x)=10/(sqrt(2))m//s`
(since there no change in the horizontal velocity)
`v_(y)^(2)=u_(y)^(2)=2(-g)(h)`
`impliesv_(y)^(2)=100/2-2xx10xx1=30`
Therefore net velocity of `CM=sqrt(v_(x)^(2)+v_(y)^(2))`
`=sqrt(100/2+30)=sqrt(80)=4sqrt(5)m//s`