Height of auditorium is 30 m and ball just reaches to this height so this is the maximum height of projectile.Let angle of projection is θ. So, Hmax=30=2gu2sin2θ30=2×10(30)2sin2θsin2θ=900600=32⇒sinθ=32 and cosθ=31Now, length of auditorium will be equal to range of projectile. So, R=gu2sin2θ=gu2(2sinθcosθ)R=0.10(30)2×2×32×31R=3018002=602m