Initial velocity of bullet will be approximately 1258.46 m/sExplanation:Given,Mass of bullet, m = 20 g = (20 / 1000) kg = 0.02 kg Initial velocity of bullet it is travelling with = V Mass of wooden BLOCK, M = 4 kg The wooden block is suspended from two light strings. Bullet on firing GET EMBEDDED into the block.  (We will assume strings massless)Height to which block-bullet combination is rised, h = 2 m To find,Initial velocity of bullet, v =?Solution,Let, the velocity of block-bullet combination just after the collision be V & the initial velocity of wooden block would be, u = 0then,By the law of conservation of linear momentum → Initial momentum before collision = Final momentum after collision→ m v + M u = m V + M V → m v + M ( 0 ) = V ( m + M )→ m v = V ( m + M )→ V = ( m v ) / ( m + M )→ V = ( 0.02 v ) / ( 0.02 + 4 )→ V = ( 0.02 v ) / 4.02    ______equation(1)Therefore, we have the the velocity of block-bullet combination after collision.Now,By the law of conservation of mechanical energy → Kinetic energy of system = Potential energy of system→ 1/2 ( m + M ) V² = ( m + M ) g h [ where g is ACCELERATION due to gravity ]→ 1/2 ( 0.02 + 4 ) V² = ( 0.02 + 4 ) ( 9.8 ) ( 2 )→ V² = 39.2Using equation (1)→ [( 0.02 v ) / 4.02]² = 39.2→ 0.0004 v² / 16.1604 = 39.2→ v = 1258.46  m/sTherefore,Initial velocity of bullet will be approximately 1258.46 m/s.