An object was launched with a velocity of 20 ms−1 at an angle of 45° to the vertical. At the
top of its trajectory the object broke into two equal pieces. One piece fell vertically
downwards. Where would the other piece fall? (Take g = 10 ms−2)
as there is no external force on the particle except the gravitation the centre of mass would go in same path as it would have gone before breaking.So after the broken pieces fall on ground their centre of mass should be at the range of the parabolic path.
Range is R=[(u^2 sin2@)/g]
Solving for @=45,u=20,g=10 we get R=40
Therefore centre of mass position is at 40
Applying conservation of moments about origin we get (m1d1+m2d2)=md----1
Here m=total mass,m1=mass of first piece(m/2),m2=mass of second piece(m/2)
d=distance of centre of mass from origin(R=40),d1=distance of first piece from origin((R/2)=20)
d2=distance of second piece from origin(we have to find)
Solving keeping these values in equation 1 we get d2=60
THEREFORE THE SECOND PIECE WILL FALL AT 60metres FROM ORIGIN