A particle P collides with `15 ms^(-1)` at an angle `60^(@)` from vertical on smooth horizontal surface. The coefficient of restitution (e) is `(1)/(sqrt(3))`. The range R, as shown in diagram will be : [Take `g =10 m//s^(2)`]

Options

A. 12.5 m
B. 12 mm
C. 11.25 m
D. 12.75 m

Answer

C

Explanation

Given `e = (1)/(sqrt3) tan phi = (1)/(e ) tan theta`
`tan phi = 3 rArr sin phi = (3)/(sqrt10)`
Along horizontal, `m xx 15 sin 60^(@) = m xx v sin phi`
`m xx 15 sin 60^(@) = m xx v sin phi`
`v = (15 sqrt5 sqrt10)/(2 xx 3) = 13.7 ms^(-1)`
`x = (v^(2) sin 2 ((pi)/(2) - phi))/(g) = (v^(2) sin 2 phi)/(g)`
`= ((13.7)^(2) xx (2 xx 3)/(10))/(10) = 11.25m`