If the angle of projection of a projector with same initial velocity exceed or fall short of `45^(@)` by equal amount `alpha`, then the ratio of horizontal rages is

Options

A. `1:2`
B. `1:3`
C. `1:4`
D. `1:1`

Answer

D

Explanation

Given,`theta_(1) = theta_(2) = 45^(@)`
`theta_(1) + theta_(2) = (45^(@) +alpha) +(45^(@) -alpha) = 90^(@)`
`u_(1) = u_(2) = u rArr R_(1) = R_(2) = ?`
`R_(1) = (u_(1)^(2)sin (theta_(1)+theta_(2)))/(g)`
`rArr R_(1) = (u^(2)sin 90^(@))/(g) = (u^(2))/(g)` ........(i)
Similarly, `R_(2) = (u_(2)^(2)sin (theta_(1)+theta_(2)))/(g)`
`rArr R_(2) = (u^(2)sin 90^(@))/(g) = (u^(2))/(g)` .....(ii)
The ratio of horizontal ranges
`(R_(1))/(R_(2)) = (u^(2)//g)/(u^(2)//g) = 1:1 rArr R_(1):R_(2) = 1:1`