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

Options

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

Answer

A

Explanation

`R = (u^(2)sin 2 theta)/(g)` at angles `theta` and `90^(@) - theta`
Now, `t_(1) = (2u sin theta)/(g)`
and `t_(2) = (2u sin (90^(@)- theta))/(g) = (2u cos theta)/(g)`
`:. t_(1)t_(2) = (2)/(g)[(u^(2)sin 2 theta)/(g)] = (2R)/(g)`