(I just noticed that you are saying that the author claims that the velocity when the ball hits the ground is -33.43 m/s. Assuming "Wind and air resistance is ignored" as you say, that is incorrect. The velocity when the ball hits the ground is -34 m/s.)

Well, yes. Since the force of gravity on an object of mass m is -9.8m (the -9.8 comes from the mass of the earth and the average distance from the center of the earth to the object- that actually varies slightly at different places on the earth) and force is mass times acceleration, ma, we have ma= -9.8m. The m's cancel leaving any object falling, under gravity, with acceleration 9.8 m/s^2 downward. Since a= dv/dt, to go from a= -9.8, integrate to get v(t)= -9.8t+ v0 where v0 is the velocity at t= 0. Since v= dx/dt, integrate again to get x(t)= -9.8(t^2/2)+ v0t+ x0= -4.9t^2+ v0t+ x0 where x0 is the x position at t= 0.

With v0= 34 m/s and x0= 2, x(t)= -4.9t^2+ 34t+ 2 as you have. "x0= 2" was the "distance off the ground" so the object hits the ground when x(t)= -4.9t^2+ 34t+ 2= 2. That, of course, simplifies to -4.9t^2+ 34t= t(-4.9t+ 34)= 0. That quadratic equation has two roots. t= 0 is, of course, when the object is initially thrown up. The other root, from -4.9t+ 34= 0, t= 34/4.9= 6.94 seconds, is the time the object hits the ground. Put that into v= -9.8t+ 34 to get the velocity when it hits the ground.

Another, simpler, way to get this result is to ignore "-4.9t^2+ 34t+ 2" and use "conservation of energy". Gravity is a "conservative" force so the kinetic energy of the object when it hits the ground must be exactly the same as the energy when it was initially thrown up. Since kinetic energy is (1/2)mv^2 and the mass is constant, the **velocity** when the ball hits the ground is exactly the same, but of opposite sign, as when it was initially thrown up!