A particle of mass 3m is projected from the ground at some angle with horizontal. Its horizontal range is R. At the highest point of its path it breaks into two pieces of masses m and 2m respectively. The smallar mass comes to rest. The larger mass finally falls at a distance x from the point of projection where x is equal to

Options

A. `(3R)/(4)`
B. `(3R)/(2)`
C. `(5R)/(4)`
D. `3R`

Answer

C

Explanation

`x_(cm)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))`
`R=(m((R)/(2))+2m(x_(2)))/(3m)rArrx_(2)=(5R)/(4)`