A projectile can have the same range R for 2 angels of projection .If t1 and t2 be the time of flight in 2 case then the product of 2 times the time of flight is oro portion al to

It is clear that A+B=90 degree

Then,

Time of flight of 1st projectile(T1) =2u sinA/g (g=acceleration due to gravity)

Time of flight of 2nd projectile(T2) =2u sinB/g=2u sin(90-A)/g=2u cosA/g

Range(R) of two projectiles are u2sin2A/g and u2sin2B/g respectively.

Putting B=90-A, ranges are proved to be equal.

Now,

T1XT2=2X2 u2sinAcosA/g2=2 u2(2sinAcosA)/g2=2 u2sin2A/g2=2R/g.

Hence, R=g(T1XT2)/2.