For a given velocity, a projectile has the same range R for two angles of projection. If `t_(1)` and `t_(2)` are the time of flight in the two cases, then `t_(1)t_(2)` is equal to

Answer

2

Explanation

`t_(1)=(2usintheta)/(g)," "t_(2)=(2u sin (90^(@)-theta))/(g)`
`t_(1).t_(2)=(2usintheta)/(g).(2u sin (90^(@)-theta))/(g)," "t_(1)t_(2)=(2xx2u^(2)sin theta cos theta)/(g^(2))`
`t_(1)t_(2)=(2u^(2))/(g).(2 sin theta cos theta)/(g)," "t_(1)t_(2)=(2u^(2))/(g).(sin 2theta)/(g)" "t_(1)t_(2)=(2R)/(g)`