Question

A projectile may have same range R for angles of projection if t1 and t2 are time of flight prove that t1.t2= 2R/g

Answer 1

projectile can have same range for angles of projection ϴ and 90-ϴ

=> T1=2Usinϴ/g T2=2Usin(90-ϴ)/g=2Ucosϴ/g

T1*T2=4U^2sinϴcosϴ/g^2=2R/g [R is range]

so it is directly proportional to range of projectile