CBSE Class 11-science Answered

Let u be the initial projection velocity and θ is initial projection angle. Let g be the acceleration due to gravity .

Maximum height h reached by projectile =

h = [ u² sin²θ ] / ( 2 g )

If angle of prjection θ = 45^o , then maximum height H reached by projectile is

H = u² / (4g)  ............................(1)

Range r of projectile is

r = ( u² / g ) sin2θ 

Maximum range R of projectile when θ = 45^o is

R = ( u² / g ) ............................( 2 )

From (1) and (2) , we get

R = 4 H