Galileo writes that for angles of projection of a projectile at angles `(45 + theta)` and `(45 - theta)`, the horizontal ranges described by the projectile are in the ratio of (if `theta le 45`)

Options

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

Answer

C

Explanation

For a projectile launched with velocity u at an angle `theta` horizontal range is given by `R=(u^(2)sin2theta)/(g)`
(I) For `theta_(1)=45^(@)-theta`
`R_(1)=(u^(2)sin(90^(@)-2theta))/(g)=(u^(2)costheta)/(g)`
(ii) For `theta_(2)=45^(@)+theta`,
`R_(2)=(u^(2)sin(90^(@)+2theta))/(g)=(u^(2)cos2theta)/(g)=R_(1):.(R_(2))/(R_(1))=1`