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The speed of the block as it leaves the table is vo / 101, the change in kinetic energy can be calculated using initial and final kinetic energies, and the distance x is derived from projectile motion equations.The question involves a collision between a bullet of mass m and a block of mass 100m on a frictionless table. The following expressions are derived:The speed of the block as it leaves the table, assuming a perfectly inelastic collision, is given by vblock = vo / 101.The change in kinetic energy (ΔKE) of the bullet-block system during impact can be calculated using KE initial - KE final, where KE initial is ½mvo² and KE final is ½(101m)(vo/101)².The distance x is derived from projectile motion equations, x = vblock * sqrt((2h)/g), where g is the acceleration due to gravity.For parts (d) and (e), where the bullet passes through the block:The time for the block to reach the floor remains the same, as it is dependent only on the height h and gravity g.The distance x for the block would now be less because the block would have a lower horizontal velocity after being struck by the bullet....