For an object thrown at 45° to the horizontal, the maximum height H and horizontal range R are related as

As we know that, maximum height of a projectile is given byHmax=u2sin2θ2gwhere, u = initial velocity of projectile            g = acceleration due to gravityand      θ = angle of projection.As per question,Hmax=v2sin24502g                  ... (i) (As, u = v, θ = 45°)Now, range of a projectile is given byR =u2 sin 2θg⇒  R =v2 sin (2 × 450)g⇒  R =v2 sin 900g              ... (ii)On dividing Eq. (i) by Eq. (ii), we getHmaxR=v2sin245°×g2g×v2sin90°=14×1⇒  R=4Hmax=4H