A projectile has same range for two angules of projection. If times of flight in two cases are ` t_(1) and t_(2)` then the range of the projectilie is

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

A projectile can have same range R for two angles of projection. It t_1 and t_2 are the times of flight in the two cases, then what is the product of two times of flight ?

A projectile can have same range R for two angles of projection. It t_1 and t_2 are the times of flight in the two cases, then what is the product of two times of flight ?

A projectile can have the same range R for two angles of projection. If t_1 and t_2 are the times of flight in the two cases, then the product of the two time of flight is proportional to A. R^2 B. frac(1)(R^2) C. frac(1)(R) D. R

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

A projectile can have the same range R for two angles of projection .If t_1 and t_2 be the times of flight in the two cases, then the product of the two times of fight is directly proportional to

A projectile has the same range R for angles of projections. If T_(1) and T_(2) be the times of fight in the two cases, then ( using theta as the angle of projection corresponding to T_(1) )

A projectile has the same range R for angles of projections. If T_(1) and T_(2) be the times of fight in the two cases, then ( using theta as the angle of projection corresponding to T_(1) )