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. R
B. `R^(2)`
C. `R^(3)`
D. `sqrt( R)`

Answer

A

Explanation

Time of flight of a projectile `=(2 u sin theta)/g`
Horizontal range of the projectile `=(u^(2) sin 2 theta)/g`
Since, the range is same, angles of projectile should be `theta` and `(90^(@) - theta)`
`t_(1) = (2u sin theta)/g, t_(2) =(2u sin (90^(@)- theta))/g = (2u cos theta)/g`
`therefore t_(1)t_(2) = ((2u sin theta xx 2u cos theta)/g^(2)) = (4u^(2) sin theta cos theta)/g^(2)`
`therefore t_(1)t_(2) = (2R)/g`, hence, `t_(1)t_(1) prop R`.