A shell is fired from a gun with a muzzle velocity u m/sec at an angle `theta` with the horizontal. At the top of the trajectory the shell explodes into two fragments P and Q of equal mass. If the speed of the fragment P immediately after explosion becomes zero, where does the centre of mass of the fragments hit the ground?

Options

A. `(u^(2)sin^(2)theta)/g`
B. `(u^(2)sin2theta)/g`
C. `(u^(2)sin^(2)theta)/(2g)`
D. `(usintheta)/g`

Answer

B

Explanation

Horizontal distance travelled by the centre of mass before hittinng the ground is `R=(u^(2)sin2theta)/g` (since the path of the centre of mass does not change due to the force of explosion)